Buy BOOKS at Discounted Price

Structure Of Atom

Class 11th Chemistry Part I Bihar Board Solution
Exercise
  1. (i) Calculate the number of electrons which will together weigh one gram. (ii) Calculate…
  2. (i) Calculate the total number of electrons present in one mole of methane. (ii) Find (a)…
  3. How many neutrons and protons are there in the following nuclei?
  4. Write the complete symbol for the atom with the given atomic number (Z) and atomic mass…
  5. Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the…
  6. Find energy of each of the photons which (i) Corresponding to light of frequency 3×10^15…
  7. Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 ×…
  8. What is the number of photons of light with a wavelength of 4000 pm that provide 1J of…
  9. A photon of wavelength 4 × 10-7 m strikes on metal surface, the work function of the metal…
  10. Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium…
  11. A 25-watt bulb emits monochromatic yellow light of wavelength of 0.57μm. Calculate the…
  12. Electrons are emitted with zero velocity from a metal surface when it is exposed to…
  13. What is the wavelength of light emitted when the electron in a hydrogen atom undergoes…
  14. How much energy is required to ionise a H atom if the electron occupies n = 5 orbit?…
  15. What is the maximum number of emission lines when the excited electron of an H atom in n =…
  16. (i) The energy associated with the first orbit in the hydrogen atom is -2.18 × 10-18 J…
  17. Calculate the wave number for the longest wavelength transition in the Balmer series of…
  18. What is the energy in joules, required to shift the electron of the hydrogen atom from the…
  19. The electron energy in hydrogen atom is given by En= (-2.18 × 10-18)/n2J. Calculate the…
  20. Calculate the wavelength of an electron moving with a velocity of 2.05 × 10^7 m s-1.…
  21. The mass of an electron is 9.1 × 10-31 kg. If its K.E. is 3.0 × 10-25 J, calculate its…
  22. Which of the following are isoelectronic species i.e., those having the same number of…
  23. (i) Write the electronic configurations of the following ions: (a) H- (b) Na+ (c) O2- (d)…
  24. What are the atomic numbers of elements whose outermost electrons are represented by (a)…
  25. Which atoms are indicated by the following configurations? (a) [He] 2s1 (b) [Ne] 3s2 3p3…
  26. What is the lowest value of n that allows g orbitals to exist?
  27. An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this…
  28. An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of…
  29. Give the number of electrons in the species and
  30. (i) An atomic orbital has n = 3. What are the possible values of l and ml? (ii) List the…
  31. Using s, p, d notations, describes the orbital with the following quantum numbers. (a)…
  32. Explain, giving reasons, which of the following sets of quantum numbers are not possible.…
  33. How many electrons in an atom may have the following quantum numbers? (a) n = 4, ms= - 1/2…
  34. Show that the circumference of the Bohr orbit for the hydrogen atom is an integral…
  35. What transition in the hydrogen spectrum would have the same wavelength as the Balmer…
  36. Calculate the energy required for the process The ionization energy for the H atom in the…
  37. If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which…
  38. 2 ×10^8 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if…
  39. The diameter of zinc atom is 2.6 Å. Calculate (a) radius of zinc atom in pm and (b) number…
  40. A certain particle carries 2.5 × 10-16C of static electric charge. Calculate the number of…
  41. In Milikan’s experiment, static electric charge on the oil drops has been obtained by…
  42. In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum…
  43. Symbols ιμγωιδτη=∀115∀ηειγητ=∀32∀σρχ=∀2−χηαπτερ_φιλεσ/ιμαγε014.πνγ∀ can be written,…
  44. An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign…
  45. An ion with mass number 37 possesses one unit of negative charge. If the ion contains…
  46. An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons…
  47. Arrange the following type of radiations in increasing order of frequency: (a) radiation…
  48. Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons…
  49. Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate…
  50. In astronomical observations, signals observed from the distant stars are generally weak.…
  51. Lifetimes of the molecules in the excited states are often measured by using pulsed…
  52. The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm.…
  53. The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and…
  54. Following results are observed when sodium metal is irradiated with different wavelengths.…
  55. The ejection of the photoelectron from the silver metal in the photoelectric effect…
  56. If the photon of the wavelength 150 pm strikes an atom and one of its inner bound…
  57. Emission transitions in the Panchen series end at orbit n = 3 and start from orbit n and…
  58. Calculate the wavelength for the emission transition if it starts from the orbit having…
  59. Dual behaviour of matter proposed by de Broglie led to the discovery of electron…
  60. Similar to electron diffraction, neutron diffraction microscope is also used for the…
  61. If the velocity of the electron in Bohr’s first orbit is 2.19 × 10^6 ms-1, calculate the…
  62. The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 ×…
  63. If the position of the electron is measured within an accuracy of + 0.002 nm, calculate…
  64. The quantum numbers of six electrons are given below. Arrange them in order of increasing…
  65. The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6…
  66. Among the following pairs of orbitals which orbital will experience the larger effective…
  67. The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will…
  68. Indicate the number of unpaired electrons in: (a) P - 1s^2 2s^2 2p^6 3s^2 3p^3 (b) Si -…
  69. (a) How many sub-shells are associated with n = 4? (b) How many electrons will be present…

Exercise
Question 1.

(i) Calculate the number of electrons which will together weigh one gram.

(ii) Calculate the mass and charge of one mole of electrons.


Answer:

(i) As we know that

Mass of one electron = 9.1× 10-28 g


1 electron weighs 9.1× 10-28 g


So, X electrons will weigh 1 g


X = [1] / [9.1× 10-28]


= 1.098×1027 electrons


Therefore 1.098×1027 electrons together weigh 1g.


(ii) By Avogadro’s Law we know that


1 mole of electron = 6.023×1023 electrons


Mass of one electron = 9.1× 10-28g


Therefore, Mass of 1 mole of electrons = [6.023×1023] × [9.1× 10-28]


= 5.4809× 10-4g


We also know that charge of one electron = 1.6× 10-19 C


Therefore, Charge of 1 mole of electron is = [6.023×1023] × [1.6× 10-19]


= 96.368×103 C


Therefore, mass of one mole of electron is 5.481× 10-4 g and the charge is 96.37×103 C.


Question 2.

(i) Calculate the total number of electrons present in one mole of methane.

(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of 14C. (Assume that mass of a neutron = 1.675 × 10–27 kg).

(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3 at STP.

(iv) Will the answer change if the temperature and pressure are changed?


Answer:

(i) As we know in 1 molecule of methane, 1 carbon atom and 4 atoms of hydrogen is present.

In carbon there are six electrons and hydrogen consist of one electron each.


So total number of electrons in methane = 6 + 4


= 10 electrons


By Avogadro’s Law we know that


1 mole of methane contains 6.023×1023 atoms.


So total number of electrons of in 1 mole methane = 10×6.023×1023


= 6.023×1024 electrons


So total number of electrons in 1 mole of methane is 6.023×1024 electrons


(ii) Mass of one neutron = 1.675 × 10–27 kg


1 mole of Carbon atom = 6.023×1023 atoms


Number of neutrons in 1 carbon atom = 14-6= 8


So total number of neutrons in 14g of Carbon = 6.023×1023×8 neutrons


So 7mg of Carbon will contain =


= [3.37288×1022]/14


= 2.4092×1021 neutrons


So, Mass of 2.4092×1021 neutrons = [2.4092×1021] × [1.675 × 10–24]


= 4.035×10–3 g


So, in 7 mg of carbon total number of neutrons is 2.41×1021 and the total mass of the neutrons is 4.035× 10–3 g


(iii) Molecular Mass of Ammonia = 17g


By Avogadro’s Law,


1 mole of Ammonia = 17g of Ammonia = 6.023×1023 atoms


Total Number of Protons in Ammonia = 7 + 3= 10


So total number of protons in Ammonia = 6.023×1024 protons


17g of Ammonia contain 6.023×1024 protons


So, 34 mg of Ammonia will contain X number of protons


X =


X = [6.023×1024] × [2×10-3]


X = 1.2046×1022 protons


Mass of one proton = 1.6726 × 10–24g


So, Mass of 1.2046×1022 protons = [1.6726 × 10–24] × [1.2046×1022]


= 20.148×10-3 g


So, in 34 mg of ammonia total number of protons is 1.205×1022 and the total mass of the protons is 20.148×10-3 g.


(iv) No, the answer will not vary with the change in temperature and pressure because the number of subatomic particles like protons, neutrons and electrons is fixed for each and every element and it does not vary with temperature and pressure.



Question 3.

How many neutrons and protons are there in the following nuclei?



Answer:

Suppose for element X with following representation

zXA


Where A = Mass Number of Element and Z = Atomic Number of the element.


(i) 6C13


Mass Number of Carbon, A = 13


Atomic Number of Carbon, Z = 6


Number of protons = Number of Electrons = Atomic Number =6


Number of neutrons = A – Z


= 13-6 = 7


(ii) 8O16


Mass Number of Oxygen, A = 16


Atomic Number of Oxygen, Z = 8


Number of protons = Number of Electrons = Atomic Number =8


Number of neutrons = A – Z


= 16-8= 8


(iii) 12Mg24


Mass Number of Magnesium, A = 24


Atomic Number of Magnesium, Z = 12


Number of protons = Number of Electrons = Atomic Number =12


Number of neutrons = A – Z


= 24-12 = 12


(iv) 26Fe56


Mass Number of Iron, A = 56


Atomic Number of Iron, Z = 26


Number of protons = Number of Electrons = Atomic Number = 26


Number of neutrons = A – Z


= 56-26 = 30


(v) 38Sr88


Mass Number of Strontium, A = 56


Atomic Number of Strontium, Z = 26


Number of protons = Number of Electrons = Atomic Number = 38


Number of neutrons = A – Z


= 88-38= 50


Question 4.

Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)

(i) Z = 17, A = 35.

(ii) Z = 92, A = 233.

(iii) Z = 4, A = 9.


Answer:

Suppose for element X with following representation

zXA


Where A = Mass Number of Element and Z = Atomic Number of the element.


(i) The representation with given atomic number and mass number is as follows:


Element with Atomic Number 17 is Chlorine.


17Cl35


(ii) The representation with given atomic number and mass number is as follows:


Element with Atomic Number 92 is Uranium.


92U233


(iii) The representation with given atomic number and mass number is as follows:


Element with Atomic Number 4 is Beryllium.


4Be9



Question 5.

Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wave number of the yellow light.


Answer:

Given:

Wavelength of Yellow Light = 580 nm.


To Find Frequency of the light:


We know the following basic relation,


Speed of Light = [Frequency] × [Wavelength]


We know speed of light = 3×108 m/s


Frequency, v = [3×108] / [580 × 10-9]


v = 5.17×1014 Hz


Wave Number is defined as the number of wavelengths which can be accommodated in one centimetre of length in the direction of propagation.


Wave Number for Yellow Light = [1] / [λ]


= [1] / [580 × 10-9]


= 1.7 × 106 m-1


Therefore, the frequency of yellow light is 5.17×1014 Hz and the wave number of yellow light is 1.7 × 106 m-1



Question 6.

Find energy of each of the photons which

(i) Corresponding to light of frequency 3×1015 Hz.

(ii) Have wavelength of 0.50 Å.


Answer:

By Planck’s Quantum Theory we have the following relation:

Energy, E = h×v


Where


h = Planck’s constant = 6.626×10-34 Js


v = frequency


Using the Planck’s relation, we will solve the numerical.


(i) Frequency, v =3×1015 Hz


By Planck’s relation we have,


Energy, E = h×v


= [6.626×10-34] × [3×1015]


= 1.9878×10-18 J


Therefore, the energy of the photon corresponding to light of frequency 3×1015 Hz is 1.988×10-18 J.


(ii) We know the following basic relation,


Speed of Light = [Frequency] × [Wavelength]


We know speed of light = 3×108 m/s


Frequency, v = [3×108] / [0.5 × 10-10]


v = 6×1018 Hz


By Planck’s relation we have,


Energy, E = h×v


= [6.626×10-34] × [6×1018]


= 3.9756×10-15 J


Therefore, the energy of the photon corresponding to light of frequency 3×1015 Hz is 3.98×10-15 J.



Question 7.

Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 × 10–10 s.


Answer:

Given:

Period, T = 2.0 × 10–10 s


Frequency, v = [1] / T


= [1] / [2.0 × 10–10]


= 5.0 × 109 Hz


We know the following basic relation,


Speed of Light = [Frequency] × [Wavelength]


We know speed of light = 3×108 m/s


Wavelength, λ = [3×108] / [5.0 × 109]


= 0.06 m


Wave Number = [1] / [λ]


= [1/0.06]


= 16. 667 m-1


Therefore, the frequency of wave is 5.0 × 109 Hz and its wavelength is 0.06 m and its wave number 16. 667 m-1



Question 8.

What is the number of photons of light with a wavelength of 4000 pm that provide 1J of energy?


Answer:

Given:

Wavelength, λ = 4000 pm


Energy, E = 1 J


By Planck’s relation we have,


Energy, E = h×v


But we know v = [c] / [λ]


Where


c = Speed of Light


v= Frequency


λ = Wavelength


So, E = hc /λ


= [[6.626×10-34] × [3×108]] / [4000×10-12]


= [1.9878×10-25] / [4000×10-12]


= 4.9695×10-17 J


4.9695×10-17 J is the Energy of 1 photon


So, 1 J is the energy of X number of photons


X = [1] / [4.9695×10-17]


X = 2.012×1016 photons


So, the number of photons which provide 1 J of energy is 2.012×1016 photons.



Question 9.

A photon of wavelength 4 × 10–7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron (1 eV= 1.6020 × 10–19 J).


Answer:

Given:

Wavelength, λ = 4 × 10–7 m


Work Function of metal = 2.13 eV


Finding Energy of Photon:


By Planck’s relation we have,


Energy, E = h×v


But we know v = [c] / [λ]


Where


c = Speed of Light


v= Frequency


λ = Wavelength


So, E = hc /λ


= [[6.626×10-34] × [3×108]] / [4×10-7]


= [1.9878×10-25] / [4×10-7]


= 4.9695×10-19 J


Therefore, the energy of the photon 4.9695×10-19 J


Finding the Kinetic Energy of Emission:


Kinetic Energy = hv - hvo


= [E – W] eV


Where


E = Energy of photon in eV


W = work function of metal in eV


= {[4.9695×10-19]/ [1.6020 × 10–19]} – 2.13


= 3.102 – 2.13


= 0.972 eV


Therefore, the kinetic energy of emission is 0.972 eV.


Finding Velocity of photoelectron:


We know the formula for kinetic energy which is given as follows:


1/2 mv2 = kinetic Energy


Where


m = mass of electron


v = velocity of electron


v2 = [2× 0.972× 1.6020 × 10–19] / [9.1× 10-31]


= [3.1143× 10–19] / [9.1× 10-31]


v2 = 3.422× 1011


Therefore v = √ [3.422× 1011]


v = 5.849× 105 m/s


Therefore, the velocity of photoelectron is 5.85× 105 m/s



Question 10.

Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol–1.


Answer:

Given:

Wavelength, λ = 242 nm


Finding Energy of Photon:


By Planck’s relation we have,


Energy, E = h×v


But we know v = [c] / [λ]


Where


c = Speed of Light


v= Frequency


λ = Wavelength


So, E = hc /λ


= [[6.626×10-34] × [3×108]] / [242×10-9]


= [1.9878×10-25] / [242×10-9]


= 8.214×10-19 J/atom


To convert the energy from J/atom to kJ mol–1 we carry out the following conversion process:


E = [8.214×10-19× 6.023×1023] / 1000


= [494.729× 103] / 1000


= 494.729 kJ/mol


Therefore, the energy of the photon 494.73 kJ/mol



Question 11.

A 25-watt bulb emits monochromatic yellow light of wavelength of 0.57μm. Calculate the rate of emission of quanta per second.


Answer:

Given:

Power of the bulb, P = 25 W


Wavelength of light emitted, λ = 0.57μm


By Planck’s relation we have,


Energy, E = h×v


But we know v = [c] / [λ]


Where


c = Speed of Light


v= Frequency


λ = Wavelength


So, E = hc /λ


= [[6.626×10-34] × [3×108]] / [0.57×10-6]


= [1.9878×10-25] / [0.57×10-6]


= 3.487×10-19 J


Rate of Emission of Quanta, R = P / E


Where


P = Power of the bulb


E = Energy of photon


R = [25] / [3.487×10-19]


= 7.169×1019 s-1


Therefore, the rate of emission of quanta is 7.17×1019 s-1.



Question 12.

Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (ν0) and work function (W0) of the metal.


Answer:

Given:

Wavelength, λ = 6800 Å


To find threshold frequency, vo:


We know the following basic relation,


Speed of Light = [Frequency] × [Wavelength]


We know speed of light = 3×108 m/s


Frequency, vo = [3×108] / [6800 × 10-10]


vo = 4.412×1014 Hz


Work Function, W0 = [6.626×10-34] × [4.412×1014]


= 2.923×10-19 J


Therefore, the threshold frequency is 4.41×1014 Hz and the work function is 2.92×10-19 J.



Question 13.

What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?


Answer:

By referring to Balmer Formula:

We know that


Wave Number


Here n1 = 2 and n2 = 4 and the value of Rh = 109678


Wave Number


= [109678×3] / 16


= 329034 / 16


= 20564.625 cm-1


We know that Wave Number = [1] / [Wavelength, λ]


Therefore, Wavelength = [1 / 20564.625]


λ = 4.863× 10-5 cm


Therefore, the wavelength of the light is 4.86× 10-5 cm



Question 14.

How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n =1 orbit).


Answer:

The expression of energy is given by,

En = - [2.18× 10-18] Z2/n2


Where,


Z = atomic number of the atom


N = principal quantum number


For ionization from n1 = 5 to n2 = ∞,


Therefore ∆E = E2-E1 = -21.8× 10-19× [1/n22-1/n12]


= 21.8× 10-19 × [1/n22-1/n12]


=21.8× 10-19× [1/52-1/∞]


=8.72× 10-20J


For ionization from 1st orbit, n1=1, n2=∞


Therefore ∆E’ = 21.8× 10-19× [1/12- 1/∞]


= 21.8× 10-19J


Now ∆E’/∆E = 21.8× 10-19/8.72× 10-20 = 25


Thus, the energy required to remove electron from 1st orbit is 25 times than the required to electron from 5th orbit.



Question 15.

What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to the ground state?


Answer:

Given:

Present State of Electron = 6


When an electron drops from nth shell to ground state, the number of lines produced is given by = {n× [n-1]} / 2


So, here in this case, n= 6


So, number of lines produced in this transition is = {6× [6-1]}/2


=15


So, the lines produce in the transition is shown as below:



Question 16.

(i) The energy associated with the first orbit in the hydrogen atom is –2.18 × 10–18 J atom–1. What is the energy associated with the fifth orbit?

(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.


Answer:

(i) Energy of an electrons = - (2.18× 10-18)/n2

Where n = principal quantum number


Now the Energy associated with the fifth orbit of hydrogen atom is


E5 = -(2.18× 10-18)/(5)2 = -2.18× 10-18/25


E5 = -8.72 × 10-20J


(ii) Radius of Bohr’s nth for hydrogen atom is given by,


Rn = (0.0529nm) / n2


For,


N = 5


R5 = (0.0529 nm) (5)2


R5 = 1.3225 nm



Question 17.

Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.


Answer:

By referring to Rydberg Formula:

We know that


Wave Number


For Balmer series, the values of n1 = 2 and the value of Rh = 109678


Wave number is inversely proportional to the wavelength of light. So for light of long wavelength, wave number has to be minimum. For minimum condition the value of n2 should be equal to 3.


Wave Number





= 15233.0556 cm-1


Therefore, the wave number for longest wavelength is 15.233× 103 cm-1.


Question 18.

What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is –2.18 × 10–11 ergs.


Answer:

Given:

Ground State Electron energy = –2.18 × 10–11 ergs


Finding Energy:


To convert energy from ergs to joules


1 ergs is equal to 10-7 Joules


So –2.18 × 10–11 ergs = –2.18 × 10–11 × 10-7


So, Ground State Electron energy = = –2.18 × 10–18 J


Energy to shift the electron from n = 1 to n= 5 state is given by following relation:


Eh = E5 – E1


The energy of hydrogen atom is given by the following equation:


En = -2n2me4Z2/n2h2


Where


m = mass of electrons


Z = atomic mass of atom


E = charge of electron


h = Planck’s constant


So, the energy required is =


=


= 2.0928 × 10–18 J


Therefore, the energy to shift the electron from n=1 to n=5 state is 2.093 × 10–18 J.


Finding Wavelength:


By Planck’s relation we have,


Energy, E = h×v


But we know v = [c] / [λ]


Where


c = Speed of Light


v= Frequency


λ = Wavelength


So E = hc /λ


λ = hc / E


= [[6.626×10-34] × [3×108]] / [2.0928 × 10–18]


= [1.9878×10-25] / [2.0928 × 10–18]


= 9.498×10-8 m


Therefore, the wavelength is 9.5×10-8 m



Question 19.

The electron energy in hydrogen atom is given by En= (–2.18 × 10–18)/n2J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?


Answer:

Given:

Electron energy in Hydrogen atom En = [(–2.18 × 10–18)/n2] J


Energy for first state, E1 = [–2.18 × 10–18]/12


= –2.18 × 10–18 J


Energy for second state, E2 = [–2.18 × 10–18]/22


= –0.5465 × 10–18 J


By Planck’s relation we have,


Energy, E = h×v


But we know v = [c] / [λ]


Where


c = Speed of Light


v= Frequency


λ = Wavelength


So, E = hc /λ


λ = hc / E


= [[6.626×10-34] × [3×108]] / [0.5465 × 10–18]


= [1.9878×10-25] / [0.5465 × 10–18]


= 3.637×10-7 m


Therefore, the wavelength is 3.64×10-7 m



Question 20.

Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 m s–1.


Answer:

Given:

Velocity of electron = 2.05 × 107 m s–1


According to the de Broglie’s Equation:


λ = h/mv


Where,


λ = wavelength of moving particle


m = mass of electron


v= velocity of particle


h= Planck’s constant [6.62× 10-34]


λ = {[6.62× 10-34] / [9.1× 10-31] × [2.05 × 107]}


= {[6.62× 10-34] / [1.8655 × 10-23]


= 3.552× 10-11 m


Therefore, the wavelength is 3.55× 10-11 m



Question 21.

The mass of an electron is 9.1 × 10–31 kg. If its K.E. is 3.0 × 10–25 J, calculate its wavelength.


Answer:

Given:

Mass of an Electron = 9.1 × 10–31 kg


Kinetic Energy, K.E. = 3.0 × 10–25 J


1/2 mv2 = kinetic Energy


Where


m = mass of electron


v = velocity of electron


v2 = [2× 3.0 × 10–25] / [9.1× 10-31]


= [6.0 × 10–25] / [9.1× 10-31]


v2 = 659.3406 × 103


Therefore v = √ [659.3406 × 103]


v = 811.99 m/s


According to the de Broglie’s Equation:


λ = h/mv


Where,


λ = wavelength of moving particle


m = mass of electron


v= velocity of particle


h= Planck’s constant [6.62× 10-34]


λ = {[6.62× 10-34] / [9.1× 10-31] × [811.99]}


= {[6.62× 10-34] / [7.389 × 10-28]


= 8.959× 10-7 m


Therefore, the wavelength is 8.96× 10-7 m



Question 22.

Which of the following are isoelectronic species i.e., those having the same number of electrons?

Na+, K+, Mg2+, Ca2+, S2–, Ar.


Answer:

Isoelectronic species are defined as those species which belong to different atoms or ions which possess same number of electrons but different magnitude of nuclear charge.

A positive charged ion denotes the loss of an electron & A negative charged ion represents the gain of an electron by a species.


1] Number of electrons in sodium [Na] = 11


Therefore, Number of electrons in sodium ion [Na+] = 10


2] Number of electrons in potassium ion [K+] = 18


3] Number of electrons in magnesium ion [Mg2+] = 10


4] Number of electrons in calcium ion [Ca2+] = 18


5] Number of electrons in sulphur [S] = 16


∴ Number of electrons in sulphur ion [S2-] = 18


6] Number of electrons in argon [Ar] = 18


Hence, the following ions are isoelectronic species:


1] Na+ and Mg2+ [10 electrons each]


2] K+, Ca2+, S2- and Ar [18 electrons each]



Question 23.

(i) Write the electronic configurations of the following ions:

(a) H (b) Na+

(c) O2–

(d) F


Answer:

The electronic configuration of ions refers to representation of electrons in the various energy shells and subshells more collectively called as orbits of the atom.


[a] H- ion


The electronic configuration of H atom is 1s1. [Atomic number = 1]


∴ Electronic configuration of H- = 1s2


[b] Na+ ion


The electronic configuration of Na atom is 1s2 2s2 2p6 3s1. [Atomic number = 11]


∴ Electronic configuration of Na+ = 1s2 2s2 2p63s0 Or 1s2 2s2 2p6


[c] O2- ion


The electronic configuration of 0 atom is 1s2 2s2 2p4. [Atomic number = 8]


∴ Electronic configuration of O2- ion = 1s2 2s2 p6


[d] F- ion


The electronic configuration of F atom is 1s2 2s2 2p5. [Atomic number = 9]


∴ Electron configuration of F- ion = 1s2 2s2 2p6



Question 24.

What are the atomic numbers of elements whose outermost electrons are represented by

(a) 3s1 (b) 2p3 and (c) 3p5?


Answer:

[a] 3s1

Completing the electron configuration of the element as 1s2 2s2 2p6 3s1.


∴ Number of electrons present in the atom of the element


= 2 + 2 + 6 + 1 = 11


∴ Atomic number of the element = 11 [sodium]


[b] 2p3


Completing the electron configuration of the element as 1s2 2s2 2p3


∴ Number of electrons present in the atom of the element = 2+ 2 + 3 = 7


∴ Atomic number of the element = 7 [nitrogen]


[c] 3p5


Completing the electron configuration of the element as 1s2 2s2 2p6 3s2 3p5

∴ Number of electrons present in the atom of the element = 2 + 2 + 6+2+5 =17


∴ Atomic number of the element = 17 [Chlorine]


Question 25.

Which atoms are indicated by the following configurations?

(a) [He] 2s1 (b) [Ne] 3s2 3p3 (c) [Ar] 4s2 3d1.


Answer:

[a] [He] 2s1

The electronic configuration of Helium is [He] 2s1 = 1s2 2s1.


∴ Atomic number of the element = 3 [lithium, a p-block element]


[b] [Ne] 3s2 3p3


The electronic configuration of Neon is [Ne] 3s2 3p3 = 1s2 2s2 2p6 3s2 3p3.


∴ Atomic number of the element = 15 [phosphorous, a p block element]


[c] [Ar] 4s2 3d1


The electronic configuration of Argon is given as [Ar] 4s2 3d1 = 1s2 2s2 2p6 3s2 3p6 4s2 3d1.


∴ Atomic number of the element = 21 [scandium, a d block element]



Question 26.

What is the lowest value of n that allows g orbitals to exist?


Answer:

Quantum number refers to the set of four special numbers which provide the entire information about electrons of a particular atom.

Thus, the letter ‘n’ represents the principle quantum number and the Azimuthal quantum number ‘l’ can have any value between to n-1.


For n = 1 [ K shell] has L = 0 [ one subshell]


For n = 2 [L shell] has L = 0, 1 [2 subshell]


For n = 3 [M shell] has L = 0, 1, 2 [3 subshell]


For n = 4 [N shell] has L = 0, 1, 2, 3, [4 subshell]


For n = 5 [O shell] has L = 0, 1, 2, 3, 4, [5 subshell]


∴ For l = 4, minimum value of n = 5



Question 27.

An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.


Answer:

‘n’ refers to the principle quantum number

‘l’ refers to the azimuthal quantum number

‘m’ refers to the magnetic quantum number

Since we know that electron is in 3rd shell.

So, n = 3

As l = n - 1, So l = 2

When l = 2 the value of m = -2, -1, 0, +1, +2


Now, for the 3d orbital:


Principal quantum number [n] = 3


Azimuthal quantum number [L] = 2


Magnetic quantum number [m] = -2, -1, 0, 1, 2


Question 28.

An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.


Answer:

(i) In an atom the number of electrons is always equal to the number of protons to balance the charge in the atom and thus an atom is electrically neutral.

So, number of protons = number of electrons = 29


(ii) The electronic configuration of the atom is given as below


1s2 2s2 2p6 3s2 3p6 4s1 3d10


This electronic configuration refers to the electronic configuration of copper.



Question 29.

Give the number of electrons in the species and


Answer:

A positive charged ion denotes the loss of an electron & A negative charged ion represents the gain of an electron by a species.

For H+2:


Number of electrons present in Hydrogen molecule = 2


So, Number of electrons present in H+2 = 2 – 1


= 1


For H2:


Number of electrons present in Hydrogen molecule = 2


For O+2:


Number of electrons present in Oxygen molecule = 8 + 8


= 16


So, Number of electrons present in O+2 = 16 - 1


= 15



Question 30.

(i) An atomic orbital has n = 3. What are the possible values of l and ml?

(ii) List the quantum numbers (ml and l) of electrons for 3d orbital.

(iii) Which of the following orbitals are possible?

1p, 2s, 2p and 3f


Answer:

(i) ‘n’ refers to the principle quantum number

‘l’ refers to the azimuthal quantum number


‘m’ refers to the magnetic quantum number


l can have any value from to n – 1.


So, for n= 3, the permissible value of l = 0, 1, 2


M can have any value from –l, -l+1 ………..0, 1 ……… l


For l = 0


m = 0


For l = 1


m = +1, 0, -1


For l = 2


m = +2, +1, 0, -1, -2


(ii) When l = 2 the value of m = -2, -1, 0, +1, +2


Now, for the 3d orbital:


Principal quantum number [n] = 3


Azimuthal quantum number [L] = 2


Magnetic quantum number [mL] = -2, -1, 0, 1, 2


(iii) For a given value of n the value of l can range from 0 to n-1
So for n =1, l = 0


Thus, 1p is not possible
For n = 2, l = 0 and 1


Therefore, 2s and 2p are possible orbitals.
For n = 3, l = 0, 1 and 2


So 3p is not possible.



Question 31.

Using s, p, d notations, describes the orbital with the following quantum numbers.

(a) n=1, l=0; (b) n = 3; l=1 (c) n = 4; l =2; (d) n=4; l=3.


Answer:

‘n’ refers to the principle quantum number

‘l’ refers to the azimuthal quantum number


‘m’ refers to the magnetic quantum number


[a] when n = 1, L = 0 [Given] The orbital is 1s. [Can have maximum of 2 electrons]


[b] For n = 3 and L = 1 the orbital is 3p. [Can have maximum of 6 electrons].


[c] For n = 4 and L = 2 the orbital is 4d. [Can have maximum of 10 electrons]


[d] For n = 4 and L = 3, the orbital is 4f. [Can have maximum of 14 electrons]



Question 32.

Explain, giving reasons, which of the following sets of quantum numbers are not possible.

(a) n = 0, l = 0, ml= 0, ms= + 1/2

(b) n = 1, l = 0, ml= 0, ms= – 1/2

(c) n = 1, l = 1, ml= 0, ms= + 1/2

(d) n = 2, l = 1, ml= 0, ms= – 1/2

(e) n = 3, l = 3, ml= –3, ms= + 1/2

(f) n = 3, l = 1, ml= 0, ms= + 1/2


Answer:


we know
N : It is the principal quantum numbers and it can take any value from 1 to n(arbitrary number)
L : It is the number of the sub-shells of the orbit can take any value from 0 to (n-1)
specifies the orientation of the orbit (ml=2×L+1) and it can take any value from -L to L
ms : It specifies spin of the atom and it can take values ms = -1/2 and 1/2

now, let's solve the options:-

(a) The given set of quantum numbers is wrong because the value of principal quantum number that is [n] cannot be zero.

(b) This set of quantum numbers is possible as n=1>0,L=(n-1)= 0 is possible,ml=0 is also possible as l=0, and ms=-1/2 is also possible.

(c) This set is not possible because for principle quantum number n = 1, l=(n-1)=0 but L=1 according to question.

(d) This set of quantum numbers is possible as n=2>0,L=2-1=1,no. of ml=2×1+1=3 which are -1,0,1,ms=-1/2 is also possible .

(e) This set is not possible because for principle quantum number n = 3, the quantum number ‘l’ =3-1=2<3(given in question).

(f) This set of quantum numbers is possible as n=3>0,l=3-1=2 can take any value from the value 0,1,2 and ml=2×l+1=-1,0,1 and ms=+1/2 is also possible .


Question 33.

How many electrons in an atom may have the following quantum numbers?

(a) n = 4, ms= – 1/2

(b) n = 3, l = 0


Answer:

(a) Total number of electrons in an atom can be found out by the following relationship

ne = 2n2


Total no of electrons when n=4, then n = 2× 42 = 32 & half of them have ms = -1/2


(b) When n = 3, L=0 means 3s orbital which can have 2 electrons.



Question 34.

Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.


Answer:

As we know that the angular momentum of an electron is given by mvr = [nh]/ 2π ……………………………………………. [1]

Where


m = mass of the electron


v = velocity of the electron


r = radius of the orbit


h = Planck’s constant


According to the de Broglie’s Equation:


λ = h/mv


Where,


λ = wavelength of moving particle


m = mass of electron


v= velocity of particle


h= Planck’s constant [6.62× 10-34]


mv = h / λ


Substituting the above value in [1]


[hr] / λ = [nh]/ 2π


2πr = nλ


Since we know that 2πr is the circumference of a circle, so it is proved in above equation that circumference of the Bohr’s orbit is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.



Question 35.

What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum?


Answer:

For an atom, wave number = 1/λ = RHZ2 [1/n12-1/n22]

For He+ spectrum Z = 2, n2=4, n1=2


Therefore [-v] = 1/λ = RH× {4× [1/22-1/42]}


= RH× {4× [1/4 – 1/16]}


= RH× {[4× 3] / 16}


= 3RH / 4


For hydrogen spectrum, wave number = 3RH/4, Z = 1


Therefore, wave number = 1/λ = RH× 1[1/n12-1/n22]


Or


RH [1/n12-1/n22] = 3RH/4


Or


1/n12 – 1/n22 = 3/4


Which can be so for n1 = 1 & n2 = 2, i.e. the transition is from n = 2 to n = 1



Question 36.

Calculate the energy required for the process



The ionization energy for the H atom in the ground state is 2.18 × 10–18 J atom–1


Answer:

Given:

Ground State Energy = 2.18 × 10–18 J atom–1


The energy of hydrogen atom is given by the following equation:


En = -2n2me4Z2/n2h2


Where


m = mass of electrons


Z = atomic mass of atom


E = charge of electron


h = Planck’s constant


For the process mentioned in the question, energy required = En– E1


= 0 – [-2n2me4 / 12h2]


= 4 × 2n2me4/h2


But we know 2n2me4/h2 = 2.18 × 10-18J


= 4 × 2.18 × 10-18


= 8.72 × 10-18J


Therefore, the energy required for the process is 8.72 × 10-18J



Question 37.

If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.


Answer:

Given:

Diameter of carbon atom = 0.15 nm


Radius of carbon atom = [0.15 / 2] = 0.075 nm


Length of line = 20 cm


Number of carbon atoms which can be place along the line is given by n = [20× 10-2] / [0.15× 10-9]


n = 1.333× 109


Therefore, the number of carbon atoms which can be place along the line is 1.333× 109.



Question 38.

2 ×108 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.


Answer:

Given:

Number of carbon atoms, n = 2 ×108


Length of the arrangement, l = 2.4 cm


Number of carbon atoms which can be place along the line is given by n = [l] / [diameter, d]


D= [l] / [n]


D = [2.4× 10-2] / [2 ×108]


D = 1.2× 10-10 m


Therefore radius = [1.2× 10-10] / 2


= 0.6× 10-10 m


Therefore, the radius of carbon atom is 0.6 Å.



Question 39.

The diameter of zinc atom is 2.6 Å. Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.


Answer:

Given:

Diameter of carbon atom = 2.6 Å


Radius of carbon atom = [2.6 / 2] = 1.3 Å


Length of line = 1.6 cm


Number of carbon atoms which can be place along the line is given by n = [1.6× 10-2] / [2.6× 10-10]


n = 61.538× 106


Therefore, the number of carbon atoms which can be place along the line is 61.538× 106.



Question 40.

A certain particle carries 2.5 × 10–16C of static electric charge. Calculate the number of electrons present in it.


Answer:

Given:

Charge on particle = 2.5 × 10–16C


Charge on electron, = ne


Where


n = number of electrons


e = 1.6× 10-19 J


n = [2.5 × 10–16] / [1.6× 10-19]


n = 1562 electrons


Therefore, the number of electrons present is 1562 electrons.



Question 41.

In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is –1.282 × 10–18C, calculate the number of electrons present on it.


Answer:

Given:

Charge present on oil drop = –1.282 × 10–18 C


Charge of one electron = 1.6× 10-19 J


Charge on oil drop, = ne


Where


n = number of electrons


e = 1.6× 10-19 J


n = [–1.282 × 10–18 C] / [1.6× 10-19]


n = 8 electrons


Therefore, the number of electrons present on oil drop is 8 electrons.



Question 42.

In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?


Answer:

In 1911, Rutherford performed alpha rays scattering experiment to demonstrate and to find the structure of atom.

Heavy atoms have a heavy nucleus carrying a great quantity of positive charge. Hence, some alpha particles are easily deflected back on hitting the nucleus. Besides a bit of alpha particles are deflected through small angles because of large positive charge on the nucleus. So with the aid of heavy nuclei atoms it was possible to understand or detect the minute deflections of alpha particles.


If light atoms are used, their nuclei will be light & moreover, they will hold a small positive charge on the nucleus. Hence, the number of particles deflected back & those deflected through some angle will be trifling. Hence deflections of alpha particles from light atoms remain unnoticeable.



Question 43.

Symbols 79Br35 and 79Br can be written, whereas symbols 35Br79 and 35Br are not acceptable. Answer briefly.


Answer:

The general convention or method of representing an element along with its atomic mass [A] and atomic number [Z] followed globally is AZX. Atomic number of an element is fixed. However, mass number is not fixed as it depends upon the isotope taken. Hence it is essential to indicate mass number.



Question 44.

An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.


Answer:

Given:

Mass Number = 81


Percentage of neutron = 31.7% more as compared protons


Let number of protons = p


Let number of neutrons= n


p + n = 81


p + {p + [31.7× p] / 100} = 81


p + 1.317p = 81


2.317p = 81


p = 81 / 2.317


p = 94.95899


p = 35


Atomic Number = 35


The element is bromine.


The atomic symbol is 8135Br.



Question 45.

An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.


Answer:

Given:

Mass Number = 37


Percentage of neutron = 11.1% more as compared electrons


Let Number of electrons = number of protons = p


Let number of protons = p


Let number of neutrons= n


p + n = 81


p + {p + [11.1× p] / 100} = 37


p + 1.111p = 37


2.111p = 37


p = 37 / 2.111


p = 17.527


p = 17


Atomic Number = 17


The element is chlorine.


The atomic symbol is 3717Cl.



Question 46.

An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.


Answer:

Given:

Mass Number = 56


Percentage of neutron = 30.4% more as compared electrons


Let Number of electrons in ion = p

As, it has 3 positive charge

Therefore, number of protons = p + 3

As, number of neutrons are 30.4 % more as compared to number of electrons.

Therefore, number of neutrons= p + 0.304 p =1.304 p

As, we all know that mass number is the sum of number of protons and neutrons.

Therefore,

p + 3 + n = 56


p + 3 + {p + [30.4× p] / 100} = 56


p + 3 + 1.304p = 56


2.304p = 53


p = 53 / 2.304


p = 23

23 is the number of electrons present in that ion.

Therefore, the number of protons = p + 3

=23 + 3


= 26


Atomic Number = 26


The element is Iron.


The atomic symbol is Fe3+.


Question 47.

Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.


Answer:

As the frequency increases the wavelength decreases. The arrangement is given as below:

Cosmic rays < X-rays < radiation from microwave ovens < amber light < radiation of FM radio



Question 48.

Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 × 1024, calculate the power of this laser.


Answer:

Given:

Wavelength, λ = 337.1 nm


Number of photons = 5.6 × 1024


Energy, E = nhv = nhc/λ


Where


n = number of photons emitted


h = Planck’s constant


c = velocity of radiation


λ = wavelength of radiation


Substituting the values in the given expression of Energy [E]:


= {[5.6× 1024] × [6.626× 10-34] × [3× 108 ms-1]} / [337.1× 10-9m]


= [1.113168] / [337.1× 10-9m]


= 3.302× 106 J


Hence, the power of the laser is 3.302× 106 J.



Question 49.

Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance travelled by this radiation in 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.


Answer:

Given:

Wavelength, λ = 616 nm = 616× 10-9m

a)

To find frequency:


Speed of Light = [Frequency] × [Wavelength]


We know speed of light = 3×108 m/s


Frequency, v = [3×108] / [616 × 10-9]


v = 4.87×1014 Hz

b)

To find distance travelled:


Velocity of the radiation = 3× 108 m/s


Therefore, distance travelled in 30s = 30× 3× 108 = 9.0× 109m


To find the energy:


By Planck’s relation we have,


Energy, E = h×v


= [6.626×10-34] × [4.87×1014]


= 3.23×10-19 J

= 2 ev.


Therefore, the energy of the photon corresponding to light of frequency 4.87×1014 Hz is 3.23×10-19 J.

d)
Number of photons or quanta =

=

=6.239 × 1018.

Question 50.

In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 × 10–18 J from the radiations of 600 nm, calculate the number of photons received by the detector.


Answer:

Given:

Energy, E = 3.15 × 10–18 J


Wavelength, λ = 600 nm


By Planck’s relation we have,


Energy, E = h×v


But we know v = [c] / [λ]


Where


c = Speed of Light


v= Frequency


λ = Wavelength


So, E = hc /λ


= [[6.626×10-34] × [3×108]] / [600×10-9]


= [1.9878×10-25] / [600×10-9]


= 3.313×10-19 J


Therefore, number of photons received = [3.15 × 10–18] / [3.313×10-19]


= 9.5


Therefore, the number of photons received cannot be in fraction


So, number of photons received is 9.



Question 51.

Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the Nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 × 1015, calculate the energy of the source.


Answer:

Given:

Period, T = 2 ns


Frequency, v = [1] / T


= [1] / [2.0 × 10–9]


= 0.5 × 109 Hz


By Planck’s relation we have,


Energy, E = n× h×v


Where


n = number of photons


h = Planck’s constant


E = [2.5× 1015] × [6.626× 10-34] × [0.5× 109]


= 8.28× 10-10 J


Therefore, the energy of the photon is 8.28× 10-10 J.



Question 52.

The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.


Answer:

Given:

Wavelength, λ1 = 589 nm


Wavelength, λ2 = 589.6 nm


To find frequency:


Speed of Light = [Frequency] × [Wavelength]


We know speed of light = 3×108 m/s


Frequency, v1 = [3×108] / [589 × 10-9]


v1 = 5.093×1014 Hz


Frequency, v2 = [3×108] / [589.6 × 10-9]


V2 = 5.088×1014 Hz


Change in Energy, ∆E = E2-E1 = h[v2-v1]


= [6.626× 10-34] {[5.093 - 5.088] × 1014}


= [6.626× 10-34] {[5× 10-3] × 1014}


= 3.313× 10-22J


Therefore, the energy difference between two excited states is 3.31× 10-22J.



Question 53.

The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.


Answer:

Given:

Work Function for Caesium, W = 1.9 eV


To find threshold frequency:


W = hvo


Where


h = Planck’s constant


vo = threshold frequency


vo = h / w


vo = [1.9× 1.602× 10-19] / [6.626× 10-34]


= [3.0438× 10-19] / [6.626× 10-34]


= 4.5937× 1014 Hz


Therefore, the frequency is 4.59× 1014 Hz


To find wavelength:


Speed of Light = [Frequency] × [Wavelength]


We know speed of light = 3×108 m/s


Wavelength, λ = [3×108] / [4.5937× 1014]


= 6.5306× 10-7 m


Therefore, the wavelength is 6.53× 10-7 m


Finding Kinetic Energy:


K.E of ejected electron = h[v-vo] = hc[1/λ – 1/λ o]


= [6.626× 3× 10-26] [1/500× 10-9 – 1/654× 10-9]


= [6.626× 3× 10-26] × {109 × [154/327000]}


= [6.626× 3× 10-26] × [468747.1289]


= 9.3177× 10-20J


Finding Velocity of photoelectron:


We know the formula for kinetic energy which is given as follows:


1/2 mv2 = kinetic Energy


Where


m = mass of electron


v = velocity of electron


v2 = [2× 9.3177× 10-20] / [9.1× 10-31]


= [1.8654× 10–19] / [9.1× 10-31]


v2 = 2.0478× 1011


Therefore v = √ [2.0478× 1011]


v = 4.525× 105 m/s


Therefore, the velocity of photoelectron is 4.525× 105 m/s



Question 54.

Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.

λ (nm) 500 450 400

v × 10–5 (cm s–1) 2.55 4.35 5.35


Answer:

Let us suppose threshold wavelength to be λ nm the kinetic energy of the radiation is given as:

h (v-vo) = 1/2 mv2


Or


hc (1/λ -1/λo) = 1/2 mv2



……………………………. [1]


Similarly, we can also write,


……………………………………. [2]


………………………………………. [3]


Dividing equation [3] and [1]





17.6070λ - 5λ = 8803.537 – 2000


λ = [6805.537] / [12.607]


λ = 539.8 nm


The wavelength is 540 nm.



Question 55.

The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.


Answer:

Given:

Work Function of metal, W = 037eV


Wavelength, λ = 256.7 nm


From the Law of conservation of energy, the energy of an incident photon [E] is equal to the sum of the work function [W] of radiation and its kinetic energy [K.E] i.e.,


Energy of incident radiation [E] = hc/λ = [6.626× 10-34] [3× 108] [256.7× 10-9]


= 7.74× 10-19 J


Since the potential applied gives the kinetic energy to the radiation, therefore K.E of the electron = 0.35Ev


Therefore, work function = 4.83 – 0.35


= 4.48 eV


Therefore, the work function of metal is 4.48 eV



Question 56.

If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 × 107 m s–1, calculate the energy with which it is bound to the nucleus.


Answer:

Given:

Wavelength, λ = 150 pm


Velocity, v = 1.5 × 107 m s–1


Energy, E = hv = hc/λ


Where


n = number of photons emitted


h = Planck’s constant


c = velocity of radiation


λ = wavelength of radiation


Substituting the values in the given expression of Energy [E]:


= {6.626× 10-34] × [3× 108 ms-1]} / [150× 10-12m]


= [1.9878× 10-28] / [337.1× 10-9m]


= 1.3252 × 10-15 J


We know the formula for kinetic energy which is given as follows:


1/2 mv2 = kinetic Energy


Where


m = mass of electron


v = velocity of electron


K.E. = 1/2 × {[9.1× 10-31] × [1.5 × 107]2


= 1.02375× 10-16 J


Energy which bounded the electron to nucleus is:


= 13.25× 10-16J-1.025× 10-16J


=12.227× 10-16 J


= [12.227× 10-16]/ [1.602× 10-19]


=7.63× 103eV


Therefore, the energy which bounded the electron to nucleus is 7.63× 103eV.



Question 57.

Emission transitions in the Panchen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 × 1015 (Hz) [1/32 – 1/n2]

Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.


Answer:

Given:

Wavelength, λ = 1285 nm


Frequency, v = 3.29 × 1015 [1/32 – 1/n2] (Hz)


Speed of Light = [Frequency] × [Wavelength]


We know speed of light = 3×108 m/s


Frequency, v = [3×108] / [1285 × 10-9]


v = 2.3346×1014 Hz


3.29 × 1015 [1/32 – 1/n2] = 2.3346×1014


[1/32 – 1/n2] = [2.3346×1014] / [3.29 × 1015]


[1/32 – 1/n2] = 0.07427


[1/9] – 0.07427 = 1/n2


1/n2 = 0.04


1/n2 = 1 / 25


n2 = 25


n = 5


The value of n is 5.



Question 58.

Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.


Answer:

Given:

Radius, r1 = 1.3225 nm


Radius, r2 = 211.6 pm


The radius of the nth orbit of hydrogen – like particles = 0.529n2/Z Å


Now r1 = 1.3225 nm or 1322.5 pm = 52.9n12 / Z


And


R2 = 211.6pm = 52.9n22/Z


Taking the ratio of r1and r2


So r1/r2=1322.5 / 211.6 = n12/n22


n12/n22 = 6.25


n1/n2 = 2.5


Therefore n2 = 2, n1 = 5


By referring to Balmer Formula:


We know that


Wave Number


Here n1 = 2 and n2 = 5 and the value of Rh = 109678


Wave Number


= [109678×21] / 100


= 2303238 / 100


= 23032.38 cm-1


We know that Wave Number = [1] / [Wavelength, λ]


Therefore, Wavelength = [1 / 23032.38]


λ = 4.3417× 10-5 cm


Therefore, the wavelength of the light is 4.34× 10-5 cm


The transition belongs to visible region.



Question 59.

Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 × 106 ms–1, calculate de Broglie wavelength associated with this electron.


Answer:

Given:

Velocity of electron, v = 1.6 × 106 ms–1


Mass of electron, m = 9.1× 10-31 kg


According to the de Broglie’s Equation:


λ = h/mv


Where,


λ = wavelength of moving particle


m = mass of electron


v= velocity of particle


h= Planck’s constant [6.62× 10-34]


λ = {[6.62× 10-34] / [9.1× 10-31] × [1.6 × 106]}


= {[6.62× 10-34] / [1.456 × 10-24]


= 4.5508× 10-10 m


Therefore, the wavelength is 455 pm



Question 60.

Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.


Answer:

Given:

Mass of electron, m = 1.675× 10-27 kg


Wavelength, λ = 800 pm


According to the de Broglie’s Equation:


λ = h/mv


Where,


λ = wavelength of moving particle


m = mass of electron


v= velocity of particle


h= Planck’s constant [6.62× 10-34]


v = {[6.62× 10-34] / [1.675× 10-27] × [800 × 10-12]}


= {[6.62× 10-34] / [1.34× 10-36]


= 494.4776 m / s


Therefore, the velocity is 494.48 m / s.



Question 61.

If the velocity of the electron in Bohr’s first orbit is 2.19 × 106 ms–1, calculate the de Broglie wavelength associated with it.


Answer:

Given:

Velocity of electron, v = 2.19 × 106 ms–1


According to the de Broglie’s Equation:


λ = h/mv


Where,


λ = wavelength of moving particle


m = mass of electron


v= velocity of particle


h= Planck’s constant [6.62× 10-34]


λ = {[6.62× 10-34] / [9.1× 10-31] × [2.19 × 106]}


= {[6.62× 10-34] / [1.9929 × 10-24]


= 3.3248× 10-10 m


Therefore, the wavelength is 332 pm



Question 62.

The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 105 ms–1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.


Answer:

Given:

Velocity of ball, v = 4.37 × 105 ms–1


Mass of ball, m = 0.1kg


According to the de Broglie’s Equation:


λ = h/mv


Where,


λ = wavelength of moving particle


m = mass of electron


v= velocity of particle


h= Planck’s constant [6.62× 10-34]


λ = {[6.62× 10-34] / [0.1] × [4.37 × 105]}


= {[6.62× 10-34] / [43.7 × 103]


= 1.516× 10-38 m


Therefore, the wavelength is 1.516× 10-38 m.



Question 63.

If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is , is there any problem in defining this value.


Answer:

By referring to Heisenberg’s uncertainty principle, we know

∆x × ∆p = h/4∏


Where,


∆x = uncertainty in position of the electron


∆p = uncertainty in momentum of the electron


∆x = 0.002nm = 2× 10-12m [given]


Finding the value of ∆p:


∆p = [h/4π] × [1 / ∆x]


∆p = [1/0.002] × {6.626× 10-34/4× [3.14]}


= [1 / 2× 10-12] × {6.626×10-34/4× 3.14}


= [5.2728×10-35] × [5×1011]


= 2.6364 × 10-23 Jsm-1


∆p = 2.637 × 10-23 kgms-1 {1 J – 1 kgms2s-1}


Actual momentum = h / [4n× 5× 10-11]


= [6.626× 10-34] / [4× 3.14× 5× 10-11]


= 1.055 × 10-24 kg m/sec



Question 64.

The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:

1. n = 4, l = 2, ml= –2 , ms = –1/2

2. n = 3, l = 2, ml= 1 , ms = +1/2

3. n = 4, l = 1, ml= 0 , ms = +1/2

4. n = 3, l = 2, ml= –2 , ms = –1/2

5. n = 3, l = 1, ml= –1 , ms = +1/2

6. n = 4, l = 1, ml= 0 , ms = +1/2


Answer:

Quantum number refers to the set of four special numbers which provide the entire information about electrons of a particular atom.

‘n’ refers to the principle quantum number


‘l’ refers to the azimuthal quantum number


‘m’ refers to the magnetic quantum number


For n = 4 and L = 2, the orbital occupied is 4d.


For n = 3 and L = 2, the orbital occupied is 3d.


For n = 4 and L = 1, the orbital occupied is 4p.


Hence, the six electrons i.e., 1, 2,3,4,5, and 6 are present in the 4d, 3d, 4p, 3d, 3p, and 4p orbital respectively.


Therefore, the increasing order of energies of the electrons is 5[3p] < 2[3d] = 4[3d] <3[4p] = 6[4p] < 1[4d].



Question 65.

The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electron in 4p orbital. Which of these electron experiences the lowest effective nuclear charge?


Answer:

Among the mentioned orbitals, the electrons present in 4p subshell experiences the lowest effective nuclear charge as these electrons are farthest away from the nucleus.



Question 66.

Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge?

(i) 2s and 3s

(ii) 4d and 4f

(iii) 3d and 3p


Answer:

Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of an atom exerted by the nucleus of the atom. The closer the orbital, the greater is the nuclear charge experienced by the electron [s] in it and the nuclear charge is inversely proportional to the distance of the electron from the nucleus.

(i) 3s is farther away from the nucleus as compared to 2s. Hence 2s will experience larger effective nuclear charge as compared to 3s.


(ii) In this case 4d will experience more nuclear charge as compared to 4f as 4d is more near to the nucleus.


(iii) 3f is farther away from the nucleus as compared to 3p. Hence 3p will experience larger effective nuclear charge as compared to 3f.



Question 67.

The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?


Answer:

Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of an atom exerted by the nucleus of the atom. The closer the orbital, the greater is the nuclear charge experienced by the electron [s] in it and the nuclear charge is inversely proportional to the distance of the electron from the nucleus.

Silicon has greater nuclear charge [+14] than aluminium [+13].


Hence the effective nuclear charge exerted on unpair 3p electron of silicon would be greater as compared to that of aluminium.



Question 68.

Indicate the number of unpaired electrons in:

(a) P - 1s2 2s2 2p6 3s2 3p3

(b) Si - 1s2 2s2 2p6 3s2 3p2

(c) Cr - 1s2 2s2 2p6 3s2 3p6 4s1 3d5

(d) Fe - 1s2 2s2 2p6 3s2 3p6 4s2 3d6

(e) Kr - 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6


Answer:

(a) Therefore, number of unpaired electron = 3

(b) Therefore, number of unpaired electron = 2 [as p orbital can have a maximum of 6 electrons]


(c) Therefore, number of unpaired electron = 6


(d)Therefore number of unpaired electron = 4


(e) Krypton has no unpaired electron as it is noble gas.



Question 69.

(a) How many sub-shells are associated with n = 4?

(b) How many electrons will be present in the sub-shells having ms value of –1/2 for n = 4?


Answer:

(a) n = 4 [Given]

For a given value of ‘n’, ‘L’ can have values from zero to [n-1].


∴ L = 0, 1, 2, 3


Thus, four sub-shells are associated with n = 4, which are s, p, d and f.


(b) Number of orbitals in the nth shell = n2


For n = 4


Number of orbitals = 16


Each orbital has one electron with ms = -1/2


Hence there will be 16 electrons with ms = -1/2