Structure Of Atom

On the basis of Rutherford's α-particles scattering experiment, the nuclear model of atom can be concluded as:

(i) The positive charge and major portion of the mass of the atom are concentrated in an extremely small region called nucleus. Such that; if atomic radii are from 10⁻¹⁰ to 10⁻¹⁰ m then nucleus should be between 10⁻¹⁵ to 10⁻¹⁵.

(ii) The nucleus is surrounded by electrons which move around the nucleus with a very high speed in circular paths called orbits. (It resembles the solar system in which the nucleus plays the role of the sun and the electrons that of revolving planets.)

(iii) Electrons and the nucleus are held together by electrostatic forces of attraction working between them.

Hence, the concept of stationary state or orbits having fixed energy cannot be concluded from Rutherford's theory.

o The filling up of electrons in corresponding orbitals decides the electronic stable ground state configuration of an atom and usually is determined by Aufbau principle (in the figure).

According to the principal “In the ground state of the atoms, the orbitals are filled according to the order of their increasing energies.” Which actually means that, lower energy orbitals get occupied by electrons before the higher ones; hence according to the effective nuclear charge and other several factors ( n , lvalue etc.) the order of energy goes like : 1s<2s<2p<3s<3p<4s<3d....so on... and for the first 3 options :(i) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s² (ii) 1s² 2s² 2p6 3s² 3p⁶ 3d⁹ 4s²

(iii) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹this order is maintained as we can see the 4s orbital is filled first (maximum capacity is 2 electrons ) and then the 3d (maximum capacity is 10 electrons)

o According to this, the electron should enter 4s orbital first and then to 3d orbital but in case of (iv) 1s² 2s² 2p6 3s² 3p⁶ 3d⁵ 4s^1, this is a disparity to the rule hence cannot be considered as ground state electronic configuration of an atom. But it will be stable as 4s and 3d orbitals are almost the same in terms of inherent energy and d⁵ i.e. a half-filled orbital has generally attained a state of stability.

In the case of 2s orbitals, the given charge cloud diagram (Fig 2.1) clearly shows a different probability density than 1s orbital. It can be seen that the probability density first decreases sharply to zero and again starts increasing.

And from the probability density graph :

We can conclude that after reaching small maxima it decreases again and approaches zero as the value of r increases further. The region where this probability density function reduces to zero are nodes which is 1 for 2s orbital which also can be seen in Fig 2.1.

hence, option (iv) The probability density of electrons for 2s orbital decreases uniformly as the distance from the nucleus increases is incorrect as the decrease of probability density does not occur uniformly.

Other 3 options (i) 1s and 2s orbitals are spherical in shape. (ii) The probability of finding the electron is maximum near the nucleus. (iii) The probability of finding the electron at a given distance is equal in all directions are all correct and they can be seen in Fig 2.1.

The cathode ray tube experiment: In this experiment which later give birth to the concept of electrons as constituent species of an atom, for testing the properties of the particles, (i)Thomson placed two oppositely-charged electric plates around the cathode ray. The cathode ray was deflected away from the negatively-charged electric plate (Cathode) and moved (particles) towards the positively-charged electrode (Anode)- Which was the clear indication of the cathode ray was composed of negatively-charged particles. (ii)Thomson also placed two magnets on either side of the cathode ray tube and observed that this magnetic field also deflected the cathode ray. Thomson repeated his experiments using different metals as electrode materials, and found that the properties of the cathode ray remained constant no matter what material originated the cathode.

The results of cathode ray discharge tube experiment, clearly shows that cathode rays consist of negatively charged particles i.e. electrons (discovered by William Crookes) and hence, the property of cathode rays (or electrons) neither depend on the material of electrodes ( therefore, option (iii) ) is correct) nor on the nature of the gas [present in cathode ray tube.

Other 2 options (i) They start from the cathode and move towards the anode and

(ii) They travel in a straight line in the absence of an external electrical or magnetic field (Fig. 4.(b)) are also correct as they are experimentally proved.

From cathode-ray tube experiment, it was evident that cathode rays consisting a the property of having negatively charged species and the deflection of the cathode ray even in presence of magnetic field and changed material remain changed makes the facts : (i) It is a negatively charged particle., (iii) It is a basic constituent of all atoms and (iv) It is a constituent of cathode rays all true.

o The mass of electron :9.109382×10^-31(mass/ Kg)=0.00054(mass/u)=0 (approx. Mass/u)

o The mass of neutron :1.674927×10^–27(mass/ Kg)=1.00867(mass/u)=1(approx. Mass/u)

Therefore, the mass of an electron is almost non-existent in compared with neutron so, (ii) The mass of an electron is equal to the mass of a neutron is not possible by any means.

According to Thomson Model of an atom(1898), an atom takes the spherical shape and the positive charge is evenly distributed all over the sphere and also the mass is assumed to be uniformly distributed all over the atom.

These positive charge and mass get neutralised by electrons which are fixed in the sphere in such a manner that it gives rise to most stable electrostatic orientations and therefore causes overall neutrality of atoms. (the plum pudding or watermelon like structure where the positive charge is the plum and the electrons are embedded in it just like seeds )

Hence other 3 options,(ii)Spectra of a hydrogen atom. (iii) Position of electrons, protons and neutrons in an atom. (iv) Stability of atom is incorrect because J.J Thomson’s atomic model fails to give an explanation for them.

Isobars are the atoms which have an equal mass number (A) but differ in atomic number (Z) i.e. the number of protons, hence (iv) is correct. For example: ¹⁴C₆ and ¹⁴N₇ where both has the same mass number 14 and different atomic number 6 for Carbon and 7 form nitrogen.

Among other 3 options (i) they have the same atomic number but different mass number indicates isotopes.

(ii) they have the same number of electrons but different number of neutrons also talks about isotopes.

(iii)they have the same number of neutrons but different number of electrons is talking about isotones.

o Just like the s orbitals, the probability density for p-orbital also passes through a minima value i.e. zero. Besides at zero and infinite distance, as the distance from the nucleus increases.

o Hence the number of radial nodes for np orbitals is n-2 (i.e. n-1-1 for the zero and infinite distance).

So, the number of radial nodes for 3p orbital is (3-2) i.e. equals to (iv) 1.

Other 3 cases(i) 3 ,(ii) 4 and (iii) 2 are not possible .

Angular nodes are the planes passing through the nucleus (origin).

For example: In case of a P_zorbital, the XY-plane is a nodal plane, in case of d_ay orbital, there are two nodal planes passing through the origin (nucleus) and bisecting the XY- plane containing z-axis.

And these angular nodes are determined by the Azimuthal quantum number ‘l’.

o For 4d orbital n=4 and l = 2

Hence 5the number of angular nodes also = l=2.

According to Heisenberg’s uncertainty principle:

It is impossible to determine the exact position and exact momentum (i.e. velocity because of momentum= mass x velocity) simultaneously of an electron.

o One of its major implication is to rule out the existence of definite paths or trajectories of electrons, because a trajectory or a definite path of an object can be determined only if we know the exact location and the velocity of that object for a certain time being so that we can predict where it will be sometimes after that.

o But in case of sub-atomic particles like electrons, it is not possible to know both at the same time hence this principle had ruled out the idea of Bohr’s orbit or any possibility of a definite trajectory for electrons.

The other 3 options (i) Pauli’s exclusion principle, (iii) Hund’s rule of maximum multiplicity and (iv) Aufbau principle are not applicable in this case.

o Sub-shells are actually sub-levels (energy levels) for an atom. The number of sub-shells is equal to the number of the principal shell i.e. the number of principal quantum number n. Also, the azimuthal quantum number l decides the orbital angular momentum ( presence of orbitals) and for a certain n, the values of l (subshells) would be 0 to (n-1)

o Hence e for 3^rd shell or orbit, there are 3 sub-shells (0,1,2) that is, 3s, 3p and 3d.

• The 3s –subshell has only one orbital.

• The 3p-subshell has 3 orbitals.(p_x ,p_y , p_z)

• The 3d-subshell has 5 orbitals(d_xy,d_yz,dxz,d_z²,d_x²_-y² )

Hence total orbitals are = 1 +3+5 = 9.

The ‘l’ or the Azimuthal quantum number itself is the magnitude of Orbital angular momentum or subsidiary quantum number. It defines the three-dimensional structure of an orbital.

For a given value of n, l can have n values ranging from 0 to n – 1.

o Other 3 quantum numbers: n, m and s are not related to orbital angular momentum.

• The principal quantum number n signifies the shell in which the electron is found, size and energy of an orbital to a large extent. It is a positive integer N=1,2,3,4.........

• The magnetic orbital quantum number or m provides the idea of the spatial distribution of an orbital with respect to co-ordinate axes. For any value of sub-shell ‘l’, m will have 2l+1 values.

• The electron spin quantum number or s is related to the spin of an electron as an electron spins around its axis
(like the earth) and s gives information of orientation of this spin of electrons.

Chlorine exists in two isotopic forms, Cl-37 and Cl-35 (isotopes are atoms of same elements with a different mass number, i.e. only no. Of neutrons are different.)

Now to calculate the average atomic masse we have to multiply atomic masses with their respective relative abundance percentage (presence) in nature.

The relative abundance of Cl³⁷ is ~24%.

The relative abundance of Cl³⁵ is ~76%.

Thus average atomic mass=

= (37 × 0.24)+(35 × 0.76)

=8.88 + 26.6

= ~ 35.5

Hence, I the ratio of Cl-37 and Cl-35 is approximately 8.88: 26.6, which approximately comes down to the ratio 1 : 3

It can also be proved by calculating Average atomic mass

As

=

=35.5, which is as given.

Hence, (c) 1 : 3 is the correct answer.

Other 3 answers (i) 1:2

(ii) 1:1 and (iv) 3;1 are all incorrect options.

o The electronic configuration of Fe³⁺ - 1s² 2s² 2p⁶ 3s² 3p⁶3d⁵

(Z of Fe = 26, that means [Ar] 3d⁶ 4s² electronic configuration and Fe³⁺ having 3 electrons less. Ar stands for Argon having 1s² 2s² 2p⁶ 3s² 3p⁶)

o Mn²⁺ also having a [Ar] 3d⁵ configuration makes Fe³⁺, Mn²⁺ isoelectronic species; having same electronic configuration. ( Z of Mn = 25 , that means [Ar] 3d⁵ 4s² electronic configuration and Mn²⁺ having 2 electrons less. Ar stands for Argon having 1s² 2s² 2p⁶ 3s² 3p⁶)

Other 3 options:(i)Cr³⁺(Z= 24 ,[Ar] 3d³),Fe³⁺ (Z=26[Ar ]3d⁵)

(iii)Fe³⁺(Z=26[Ar]3d⁵,Co³⁺ (Z= 27 ,[Ar] 3d⁶) and

(iv) Sc³⁺ ( Z=21,[ Ar]3d⁰ ) , Cr³⁺ (Z=24 ,[Ar] 3d³),cannotbe considered to have same electronic configurations.

o Spin quantum numbers or m_s can only take 3 values, i.e. + 1/2and – 1/2which are the 2 spin states ( more than two spin states cannot be held in the orbital). And for an orbital to hold two electrons each of the 2 electrons should be of same quantum numbers i.e. m_s and opposite spins thus, cancelling each other.

o In the case of multi-electron atoms (O), Z_eff is the effective nuclear charge which is the net positive charge of the nucleus, faced by the outer shell (valence) electrons.

• The value of this Z_eff depends on the shielding or screening effect of certain orbital electrons.

• Due to the electrons in the inner shells (nearest to the core), the electron in the outer shell is unable to experience the full positive charge of the nucleus (Z_e).

• Hence, the concept effective nuclear charge is introduced which will also be lowered due to the partial screening or shielding of the positive charge on the nucleus by the inner shell electrons.

Hence the other 3 options -

(i) Z_eff for an electron in a 2s orbital is the same as Z_eff for an electron in a 2p orbital.

(cannot be because 2s and 2p orbital electrons have different screening effect and Z_eff of 2s orbital is > than 2p orbital as 2s is closer to the nucleus.)

(ii)An electron in the 2s orbital has the same energy as an electron in the 2p orbital. (in case of oxygen atom this is not possible as energy of 2p orbital is > 2s orbital.

(iii) Z_eff for an electron in 1s orbital is the same as Z_eff for an electron in a 2s orbital. (also due to the screening effect Z_eff of 1s with no screening is > than 2s ) are all incorrect.

o According to de-Broglie, where is the wavelength of a material particle, m is the mass and v is the velocity of the particle having dual nature, his plank’s constant.

o Hence, the wavelength is directly related to the mass of a particle and the relation is inversely proportional; i.e. greater the mass shorter will be the wavelength. And amongst the given particles the alpha particle has the greatest mass, therefore, it will have the shortest wavelength.

Other 3 options (i) Electron (iii) Neutron

(iv) Proton are incorrect.

Isotopes are species of same elements having different mass number (proton+neutron) number, they have the same atomic number or proton numbers but differ in numbers of neutrons.

Therefore, the entities (iii)¹⁴₆ X, ¹⁴₇ Y and (iv)⁸₄X, ⁸₅Y are not at all isotopes as they have atomic numbers 6 and 7, 4 and 5 respectively.

Other 2 options (i) ¹²₆ X, ¹³₆ y (ii) ³⁵₁₇X, ³⁷₁₇Y are definitely isotopes.

o Degenerate orbital means orbital having same energies. And the energy of an electron in the multi-electron system of atoms depends on the principal quantum number (which is the main shell,n) and the azimuthal quantum number(which is the sub-shell, l ) as well.

o Hence , electrons which are present in the same main shell and the same sub-shells as well will have the same energies and this means they should have equal n and l values too, which is the case for

(i) (a) n = 3, l = 2, m_i = –2, m_s = - 1/2and (b) n = 3, l = 2, m_i = –1, m_s= − 1/2and

(iv) (a) n = 3, l = 2, m_I = +2, m_s = − 1/2,(b) n = 3, l = 2, m_I = +2, m_s = + 1/2as well where n and l values are equal.

The other 2 options (ii) (a) n = 3, l = 1, m_i = 1, m_s = +1/2(b) n = 3, l = 2, m_l = 1, m_s = + 1/2

And (iii) (a) n = 4, l = 1, m_I = 1, m_s = + 1/2(b) n = 3, l = 2, m_I = 1, m_s = + 1/2are in correct because the electrons present in there are not having degenerate orbitals.

o The principal quantum number n is allowed to have l (azimuthal quantum numbers) values from 0 up to (n-1), i.e.1 number less than the value of n which is present in the two options -

(ii)2, 1, +1 and (iii) 3, 2, –2 where both the l (sub-shell) values are less than the n values(main shell) and magnetic quantum number m_l values are also justified ( as -l to +l values are allowed).

The other 2 options (i) 1 1 +2 and (iv) 3 4 –2

incorrect because here the l values are written equal and greater than the n values respectively.

o Electronic configurations of Na⁺ - 1s² 2s² 2p⁶ (less in one electron Na - [Ne] 3s¹ ) , therefore having 10 electrons

which resembles both Mg²⁺ - 1s² 2s² 2p⁶ (less in 2 electrons , Mg – [Ne] 3s²) and

O²– - 1s² 2s² 2p⁶ (gain in 2 electrons , O – [He] 2s² 2p⁴ ) both having equal 10 electrons .

Hence , both (i) Na⁺, Mg²⁺ and iii) Na⁺ , O²– are example of isoelectronic electron pairs.

^o Among other 2 options , (ii) Al³+ ([Ne] less in 3 electrons having total 10 electrons ), O– ( [He] 2s² 2p⁵ having total 9 electrons) and

(iv)N^3- ( [He] 2s² 2p⁶ a total of 10 electrons ), Cl^- ([Ne] 3s² 3p⁶ a total of 18 electrons) they are incorrect options.

o Azimuthal quantum number or l is the orbital angular momentum value of a certain orbital. And it defines the 3-D shape of an orbital by introducing the concept of angular momentum.

Spin quantum number m_s is the value of spin angular momentum of an electron as electron rotates around its axis (also revolves around nucleus, like earth revolving sun ) , therefore beside of having charge and mass an electron also has a certain orientation of the spin relative to the chosen axis which gives rise to the concept of spin angular quantum number (m_s).

Other 2 options

(ii) The principal quantum number determines the orientation and energy of the orbital (principal quantum number only can determine the size and energy to a large extent of the orbital, not orientation because the spin quantum number does that ) and

(iii) The magnetic quantum number determines the size of the orbital (ml determines the spatial orientation of an orbital among the chosen coordinate axes) are wrong.

The increasing order of effective nuclear charge (Z_eff) experienced by the electron present in the orbitals is d<p<s.

This is due to the shielding or screening effect t of inner electrons ( nearest to the core, s ) towards the outer shell electrons. S orbital is the nearest to the nucleus so it will experience almost the full positive charge (Ze) of the nucleus; whereas p and d shows decrease in effective nuclear charge because they are more shielded by the inner core electrons, therefore, feel less positive charge attraction by the nucleus and as d orbitals electrons are at more distance than p orbitals the electrons present in it experience less effective nuclear charge .

An oxygen atom has only 8 numbers of electrons, hence its electronic configuration is :

[He] 2s² 2p⁴ ( [He] = 1s²)

The orbital diagram of oxygen will be -

One Ni atom has 28 electrons and its electronic configuration is : [Ar] 4s² 3d⁸

It becomes Ni²⁺ by losing 2 electrons, hence configuration of Ni2+ is : [Ar] 4s⁰ 3d⁸

So, nickel loses two electrons from the 4s orbital not the 3d orbital.

• This is due to the two contrary facts: as 4s has a lower energy level than 3d orbital (Aufbau principle) then it should have been the 3d electrons that should get lost.

• But, as the 4s orbital is further from the nucleus and is shielded by the 3d electrons it is easier to lost 4s electrons than 3d electrons. Moreover, it leaves the 3 d electrons as valence shell electrons which is more stable than losing them.

The energy of orbitals depends on the principal quantum number or the main shell to a large extent. Hence, orbitals with an equal value of n will have the same levels of energy and will be called degenerate orbitals.

Amongst the given orbitals the degenerate orbitals will be 3d_xy, 3d_z2, 3d_yx because they have the same main shell n = 3.

And 4dxy_, 4d_yx, 4d_zz because they have the same value of n=4.

Nodes are the region present among the orbitals where the probability density of finding electrons will be zero.

In case of np orbitals , radial nodes = n – l – 1 = 3 –1 - 1 = 1

Angular nodes = l = 1.

(a) the increasing order of energy of the given orbital is : 1s >2s >2p> 3s

(n+l) value of 1s orbital – 1 + 0 = 1

(n+l) value of 2s orbital – 2 + 0 = 2

(n+l) value of 2p orbital – 2+ 1 = 3

(n+l) value 3s orbital – 3 + 0 = 3, among 2p and 3s, 2p has lower value of n therefore it has lower energy than 3s.

(b) the increasing order of energy of the given orbital is : 3s<3p<4s<4d

(n+l) value of 3s: 3 + 0 = 3

(n+l) valueof 3p: 3+1 = 4

(n+l) value of 4s: 4 + 0 = 4

(n+l) value of 4d: 4 + 2 = 6

Among 3p and 4s both has same n+l value but 3p has lower n value hence, 3p<4s.

(c)the increasing order of energy of the given orbital is : 4d<5p<6s<4f<5d

(n+l) value of 4d : 4 + 2 = 6

(n+l) value of 4f : 4 + 3 = 7

(n+l) value 5p : 5 + 1 = 6

(n+l) value of 5d : 5 + 2 = 7

(n+l) value of 6s = 6 + 0 = 6

Among 4d, 5p and 6s the n value is greater in 5p and 6s which is the case for 4f and 5d also therefore despite of having equal (n + l )value .

(d)the increasing order of energy of the given orbital is: 7s<5f<6d<7p

(n+l) value of 5f : 5 + 3 = 8

(n+l) value of 6d : 6 + 2 = 8

(n+l) value of 7s : 7 + 0 = 7

(n+l) value of 7p : 7 + 1 = 8 , among 5f, 7s and 7p the energy order is in accordance with n values.

II. (a) among the orbitals, 5s has the lowest energy.

the (n+l) value for 5s is the lowest = 5 + 0 = 5. Other orbitals have (n+l )value more than 5 –

5p= 5 + 1 = 6 , 4f = 4 + 3 = 7 , 4d = 4 + 2 = 6.

(b) among the orbitals , 5f has the highest energy because the (n +l ) value - 5 + 3 = 8 is highest.

5d = 5 + 2 = 7 , 5p = 5 + 1= 6 , 6s =6 + 0 = 6 , 6p =6 + 1 = 7 .

Neutron will not show deflection from the path on passing through an electric field.

This is due to the neutral nature of the neutron particles. Therefore, it has no charge and does not get affected by any electric field.

Among other 3 particles proton (positive ), electron (negative), cathode rays (beam of electrons, negatively charged) all have charges in them so they will get deflected easily by an electric field

The mass number of an atom = number of protons + number of neutrons

Therefore atomic number ( number of protons ) = mass number – no. Of neutrons.

The atomic number of atom: 13 – 7 = 6.

According to Planck’s quantum theory, energy is related to the frequency of radiation by :

E = h × Frequency

=

So, E1/λ

Hence, the relation b/w energy and wavelength are inversely proportional, therefore lesser the wavelength higher will be the energy of the radiation.

For the given wavelengths

λ(A) = 300 nm = 300 x 10^-9 m = 3 x 10 ^-7 m

λ(B) = 300 μm = 3 00 x 10^-6 = 3 x 10^-4

λ(c) = 3 nm = 3 x 10 ^{– 9}

λ (D) 30 A° = 3 X 10 ^{- 9}

the increasing order of the given wavelengths : λ(c) = λ (D) <λ(A)< λ(B)

hence the increasing order of energy will be the opposite: λ(B)< λ(A:<λ(c) = λ (D)

As 3d and 4s orbital – these 2 levels are very close in energy there are the frequent exchange of electrons occurs between them and this gives rise to the more stable configuration 3d¹⁰4s¹ for Cu than the alternative 3d⁹4s²

Two major factors are responsible for this :

1. The n+l value of 4s is 4+0 = 4 and also for 3d is 3 + 2 = 5.

• Hence, the 3d orbital is higher energy than 4s so the electron should enter 4s first and after filling up then should enter 3d but this is not the actual case (Aufbau principle)

• . as the screening effect of outside the argon, the core is not sufficient to shield the 3d orbital electrons, therefore it lacks in energy and the electrons first enter the 3d instead of 4s and 4s become the outermost valence shell and even loss of electrons occur from it.

2. Another important factor is the extra stability of a full filed (d¹⁰) configuration. There are two major reasons behind this :

i. The symmetrical distribution of the electrons paired in the d orbital.

ii. And the exchange energy of the electrons is also maximum while holding the full-filled d orbital.

o According to Bohr’s model for the hydrogen atom

_Hcm^-1

here, n₁ = 2 and n₂ = 4 and _H = Rydberg’s constant = 109677

Hence, wave number ^-v= 109677 ( )

According to de-Broglie,

λ = h / mv ,

where λ is the wavelength of the particle-wave, m and v are the mass and velocity of the particle respectively and h = Planck’s constant = 6.626 X 10^-34 J.s

hence, in this case, λ =

now, m = 100g = 0.1 kg and

v=100km/h = 100 X1000/ 60X60 = 1000/36 m/s

λ =

= 238.5 X 10^-36 m^-1

The range of wavelength is very small here (10 ^-36), therefore the wave nature of the ball cannot be detected.

The atomic spectra of hydrogen revealed that more than 4 lines coming closer together in the ultraviolet end; they converged and soon gives rise to the concept of spectral lines (instead of a definite spectrum )

In order to get the atomic spectrum of gaseous hydrogen, a discharge tube containing the gas is usually employed; I the high voltage applied a discharge is started and the molecules split into atoms. Images of the slit are formed by each resolved component of the radiation which appears as discrete illuminated lines on a dark background when viewed through a telescope.

In 1885, Balmer showed on the basis of this experimental observation, that line spectrum of the visible lines can be expressed in terms of wavenumber ( =1/λ ) and the formula will be :

Where is the Rydberg’s constant also known as R_H

Series of line described buying this formula is called Balmer series which only appears the line in the visible region of the electromagnetic spectrum.

Besides the Balmer series, in the visible region, the spectrum of hydrogen also consists of the Lyman series in the ultraviolet region and 3 others in the infrared region – Paschen, Brackett and Pfund

.

Soon it was concluded that the wavelengths of the line spectra in all series of the spectrum can be expressed as the simple relation :

,

where n_2>n₁ .

The bright-line spectrum shows that the energy levels in an atom are quantized. These lines are obtained as a result of electronic transitions between the energy .and if the electronic energy levels were continuous and not quantized or discrete; the atomic spectra would have shown a continuous absorption(from lower to higher energy level transition) or emission (from higher to lower energy level transition.

Out of electron and proton, being the lighter particle electron will have a higher velocity and will also produce matter waves of the same wavelength

According to de-Broglie in case of wave-particle duality the wavelength can be expressed as :

This λ= h/p= h/mv ,

h is the Planck’s constant and m and v be the mass and velocity of the wave particles.

By comparing both proton and neutron, the mass of the electron (m-= 9.1 X 10 -31 ) is negligible to proton (m= 1.67 X 10-27 ) hence, the value of the wavelength of the proton will be negligible in compared with the electron as λ α 1/mv

Therefore proton will not produce matter-wave.

Wavelength can be defined as the distance between two alike successive points in a wave (usually b/w two maxima s i.e. peaks or two minima s i.e. troughs shown in the fig.)

So, for the given hypothetical wave, wavelength λ = 4 X 2.16 pm

= 8.64 pm.

Relation b/w wavelength and frequency can be expressed as :

λ = c/ϑ, where c be the velocity of light and ϑ is the frequency of the radiation.

For the given problem λ = 3 X 10⁸ ms-1 / 4.620 X 10¹⁴ Hz

= 0.6494 times10^-6m^-1

The concept of the orbit is a bit ancient compared with orbital.

The orbit stands for a definite circular path for the electrons to revolve around the nucleus. It represents the two-dimensional motion of the electrons around the nucleus.

Whereas orbital is not that well-defined path because it’s a region around the nucleus where the probability of finding an electron is maximum. It gives the three-dimensional space idea for the motion of electrons.

According to Heisenberg’s uncertainty principle :

“It is fundamentally impossible to determine accurately both the velocity and the position of a particle at the same time.”

Hence ,

Where

From the given problem,

mass of the ball = 4 g and speed is = 90 m /s

hence,the uncertainity of speed is ∆v = 4/100 × 90 = 3.6 m/s

∆x is given by ∆x = h/4πm∆v

Hence , the uncertainity of postion is ∆x = 6.26 × 10^-34 / 4 × 3.14 × 4× 3.6

= 1.46 x 10^-33 m

The uncertainty principle is only significantly applicable for microscopic particles and not macroscopic particles this can be concluded from the measurement of uncertainty:

, as the uncertainty is in inverse proportion with the mass we can say that greater mass will bring smaller in insignificant uncertainty for macroscopic particles.

For example, if we take a particle or an object of mass 1 milligram i.e. 10^-6 kg )

We calculate the,

= 10^-28 m^-2 s ^-1

The value we got is negligible and very insignificant for the uncertainty principle to be applicable to the particle.

Unlike in the hydrogen atoms where the energy of an electron only depends upon the principal quantum number or the main shell, in case of multi-electron atoms the energy depends on another factor apart from n i.e. the value of l or orbital angular momentum or the sub-shell. And the energy of sub-shells for the same principal quantum number or the main shell increases in the order of s<p<d<f. And for the higher orbitals, this difference is pronounced to a larger extent which results in: 4s<3d or 6s<5d or 4s< 6p.

The main reason behind these differences is the mutual repulsion in multi-electron atoms which is absent in case of hydrogen. These repulsion besides the presence of interaction between electron and nucleus indicates that in the case of multi-electron atoms the stability comes for the occurrence of more powerful total attractive interactions over the repulsions.

Matching the following species with their corresponding ground state electronic configuration :

(i)Cu - (c) 1 S² 2s² 2P⁶ 3s² 3p⁶ 3d¹⁰ 4s¹

(ii) Cu^{2+ -}- (d) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹

(iii)Zn²⁺ - (a) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰

(iv)Cr³⁺ - (e) 1 s² 2s² 2p⁶ 3s² 3p⁶ 3d³

Matching the quantum numbers with the information provided-

(i) Principal quantum number - (b) energy and size of orbital

(ii) Azimuthal quantum number - (d) shape of the orbital

(iii)Magnetic quantum number - (a) orientation of the orbital

(iv) Spin quantum number - (c) spin of an electron

Matching the following rules with their statements:

Hund’s Rule – (c) Pairing of electrons in the orbitals belonging to the same subshell does not take place until each orbital is singly occupied.

And also

(b) Half-filled and completely filled orbitals have extra stability.

(ii) Aufbau Principle –(e) In the ground state of atoms, orbitals are filled in the order of their increasing energies.

(iii) Pauli Exclusion Principle -(a) No two electrons in an atom can have the same set of four quantum numbers.

(iv) Heisenberg’s Uncertainty Principle -(d) It is impossible to determine the exact position and exact momentum of a subatomic particle simultaneously.

(i) X-rays-(d) v = 10¹⁸ Hz

(ii) UV-(c) v = 10¹⁶ Hz

(iii) Long radio waves-(a) v = 10⁰ – 10⁴ Hz

(iv) Microwave -(b) v = 10¹⁰ Hz

Matching the following:

• (i) Photon -(d) Exhibits both momentum and wavelength

(ii) Electron -(a) Value is 4 for N

(iii) ψ²- (b) Probability density

(iv) Principal quantum number n -(c) Always positive value

matching species given in Column I with the electronic configuration given in Column II

(i)Cr -(c)[Ar]3d⁵4s^-1

(ii)Fe²⁺ -(b)[Ar]3d⁶ 4s⁰

(iii)Ni²⁺- (a)[Ar]3d⁸ 4s⁰

(iv) Cu-(d)[Ar]3d¹⁰4s¹ .

Electrons usually take part in chemical reactions and also responsible for certain chemical behaviour of an element because they are determined by the numbers of protons in the nucleus.

When it comes to isotopes they show resemblance in chemical nature and behaviours to other substances.

This is due to an equal number of protons i.e. also equal electrons present in then which controls their chemical properties.

Hence, the given Assertion (A) is correct and Reason (R) can explain it properly (i) Both A and R are true and R is the correct explanation of A. Is the right answer.

Other 3 options (ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false are incorrect.

The name “blackbody” is usually given to a body because it not only absorbs radiations of all frequencies but emits the same also.

The major factors working behind this are i. The temperature (a blackbody acts as the best emitter at thermal equilibrium) and ii. On the emissivity.

Hence, the Reason (R) - The frequency of radiation emitted by a body goes from a lower frequency to higher frequency with an increase in temperature cannot explain the assertion (A): a Black body is an ideal body that emits and absorbs radiations of all frequencies.

Although (R) being a true fact because the emission radiated by a body always depends on temperatures in accordance with Arrhenius’s law of activation energy which suggests that as the temperature increases so as the rate of radiation.

as the graph shows, the intensity of radiation increases for increasing temperatures and for shorter wavelengths ( higher frequencies).

Assertion (A) It is impossible to determine the exact position and exact momentum of an electron simultaneously is the statement of Heisenberg’s uncertainty principle which is mathematically expressed as :

And that is why a clear definite path or trajectory of an electron cannot be predicted accurately and therefore gives rise to the concept of probability of finding the electron in a certain region (probability density).

Hence, Reason (R): The path of an electron in an atom is clearly defined. Is a false statement and therefore cannot justify (A).

• When light radiation of an appropriate frequency (threshold frequency() falls on or hits a metal surface, the ejection of electrons from the metal surface takes place.

This ejection of electrons from the surface of a metal by radiations of suitable frequency is called the photoelectric effect (and the ejected electrons are called photoelectrons).

• The observations of the photoelectric effect experiment which could not be explained on the basis of classical physics (wave theory) theories are :

(i) First of all the kinetic energy possessed by the ejected electrons is directly proportional to the frequency of incident light and surprisingly is totally independent of the intensity of the incident light.

(ii) No electrons will be ejected from the metal surface unless and until the frequency of light used is greater than a specified minimum value i.e. threshold frequency (.

• The classical wave theory fails to explain these phenomenon because classical physics tells that energy of light is proportional to the intensity of light, therefore by increasing intensity solely, it should have been possible to kick-start the emission of electrons and the energy (basically kinetic here) should have been proportional to the intensity of light

• But the actual observations contradict the classical point of view in this sense that the energy of the electrons does not even depend on the intensity, rather it depends on frequency.

• Photoelectric effect can be explained by the quantum concept of radiation given by Max Planck...

The observations of photoelectric effect experimentally provide with the following conclusions –

(i) The photocurrent is proportional to the intensity of incident radiation.

(ii) The magnitude of stopping potential and hence the maximum kinetic energy of emitted photoelectrons is proportional to the frequency of the emitted radiation.

(iii) There exists a minimum threshold frequency so that if radiation of frequency lesser than this threshold frequency is incident on the metal surface, there is no photoemission irrespective of the intensity of radiation.

• Einstein explained (1905) the photoelectric effect on the basis of quantum theory, where the concept of quantized packets of energy i.e. photons (present in the light radiation and which are responsible for the emission of the electrons also )is introduced.

E=hν, where E is the energy of the incident photon (light radiation)
where
h is Plancks’s constant and ν is the frequency of radiation.

He proposed, that when a photon striking the metal surface it is using its binding energy (or (the energy which binds the electron to the nucleus) to eject the electron from the metal.

And the residual part of the energy is being transferred to the electron in the form of kinetic energy.

Therefore, the total energy of photon = binding energy + KE of the ejected electron

(v= velocity of the ejected electron)

or,, where, v₀ is the threshold frequency.

hence, kinetic energy of the ejected electron = h(v-v₀)

this expression explains the experimental facts :

• If v> v_0, the excess energy hv-hv₀is transferred to the ejected electrons as kinetic energy.

Thus, with the increase in the frequency of the incident photon, the KE of ejected electron increases (Fig)

• If, v< v0, no electrons will be emitted whatever the intensity of the radiation is.

• There will be one electron ejected for each photon, which means increasing the intensity of light of o given frequency (> v₀ ) number of photons striking the metal surface also increases., so as the numbers of electrons ejected. But the KE of electrons will be unchanged as long as a certain frequency.

• According to ,quantum theory:

hv =hv₀ + 1/2m v²

1/2m v² = h(v-v₀)

where h = plank’s constant = 6.626 x 10^-34 J S,

v₀ = threshold frequency of photons , which we have to calculate

v = frequency of incident photons. Given = 1.0×10¹⁵ s^-1

1/2m v² = kinetic energy of electrons given = 1.988 × 10^-19 J

= 1.0×10¹⁵ s^-1 - 0.300030 × 10¹⁵

= 6.9997 × 10 ¹⁴ s^-1 (threshold frequency)

• The relation b/w wavelength and frequency , v = c/λ , where λ = wavelength and c= velocity of light = 3 × 10⁸ m/s ,

• Here in case of threshold frequency v₀ = c/λ₀

Hence the maximum wavelength of the photon will be

4.36 × 10^-7 m which is definitely greater than

Now, 600 nm = 6 × 10^-7m is definitely greater than the calculated λ₀ value.

hence, the electron will not be emitted if a photon with a wavelength equal to 600 nm hits the metal surface.

When an electric discharge is passed through hydrogen gas, the hydrogen molecules dissociate to produce excited hydrogen atoms. These excited atoms emit electromagnetic radiation of discrete frequencies which can be given by the general formula :

- this interpretation of the spectrum of atomic hydrogen can be reached by using 2 basic postulates of Bohr’s model of an atom :

1. The criterion of selecting stationary orbits by the electrons follows the principle: “ the angular momentum of the electron should be an integral multiple of h/2π (h= Planck’s constant)

if the mass and velocity of the electron is m and v at an orbit of radius r then its angular momentum

mvr = n. h/2π,n= non zero positive integer, i.e. 1,2,3,..................... etc.

2. When an electron jumps from one orbit to another, the difference of energy between the two energy levels is either emitted or absorbed (in accordance with the quantum theory of radiation).

Thus, when an electron jumps from an orbit with energy E₂ to an orbit of energy E₁ (E₂>E₁₎, the difference in energy (E₂-E₁ ) is emitted in the form of quantized radiation. if the frequency of this radiation is ϑ then energy is given by :

E₂ – E₁ = hϑ

Now using these 2 points we can reach to the Rydberg’s constant –

Let us consider 2 electron orbits with quantum numbers n1 and n2 in a Bohr type (one electron system ) orbit such that n2>n1 and the corresponding energies are En1 and En2 and En1<En2.

Hence the energy emitted while transfer of electrons from n2 to n1 orbit :

E₂-E₁= hϑ, ϑ being the frequency of radiation.

According to the Bohr atomic model for hydrogen atom (Z=1):

Therefore : E2-E1 = hϑ =

Or, ϑ= and , c is the velocity of light.

Substituting the values form, e, ε₀ and h

cm^-1

Electronic transitions in the hydrogen atom are given by the Rydberg’s equation:

Where is the Rydberg constant.

And n_i and n_f indicates the initial and final energy levels (n being the principal quantum number).

• For the given problem, n_i implies to 3 and n_f implies to 2

Hence, =

= × 0. 1389

= - 3.02802 × 10^-19 J/H atom.

Here, energy is released or emitted.

When it comes to the frequency of the photon emitted, the energy would be taken in terms magnitude only,

, where, h= 6.626 × 10^-34 J S, Planck’s constant and is the frequency of the radiation (energy) emitted (photon) due to the transition of the electron.

Hence,

=

=4.569 × 10^{14 S-1}.

Although Bohr’s atomic model was undoubtedly an improvisation to Rutherford’s nuclear model but it has some major drawbacks and failure when it comes to complex structures of an atom as it was too simple.

There are a few shortcomings of this model –

• It fails to explain the multiple thin and finer lines (doublet) that are found in the spectrum of hydrogen atoms which is obtained by using advanced and elaborated spectroscopic techniques.

• It is also unable to explain the spectrum of multi-electron atoms . for example even He atom which possesses 2 electrons only.

• Another fact that cannot be derived from the model is the splitting (separation) of the spectral lines on the application of a magnetic field (Zeeman effect) or in presence of an electric field (Stark effect).

• It also could not explain the cause of chemical combination (bonding) of different atoms and shapes of molecules arising from it.

• It assumes a definite knowledge about position and momentum of electrons at the same time also declining the wave-particle dual nature of an electron. ( contradicts Heisenberg’s uncertainty principle)

That is why a change and improvisation of this model was required effectively.

When de Broglie's hypothesis and Heisenberg’s uncertainty principle came to the picture the concept of an electron moving around a defined and specific orbit became insufficient in order to describe the structure of an atom.

(i)de Broglie"s hypothesis states that every moving body posses some wave property with it which are called de Broglie waves.

(ii)Heisenberg"s uncertainty principle states that it is impossible to determine the exact position and momentum of an electron simultaneously which also contradicts orbit concept.

These 2 developed concepts of movement of an electron changed the scenario of the model where an electron moves in an orbit and later was replaced by the concept of orbital which considered a space around nucleus where the probability of finding electron is maximum.

• For the wave-like properties of the electron, it cannot be considered only as a particle and obtaining a definite position and velocity was also impossible.

• Hence electron now in the developed concepts was considered as electromagnetic energies which have quantized energy at certain energy levels which is the direct application of its wave nature.

• Again, Both the exact position and exact velocity of an electron(or any particle) in an atom cannot be determined simultaneously(at a certain time: Heisenberg’s uncertainty principle). Therefore the determination of the path of movement of an electron in an atom is almost impossible or it cannot be known accurately. That is why, , the concept of probability of finding the electron at different points in an atom comes to the picture.

The name given to the changed model of atom is “Quantum Mechanical Model of Atom” and was developed by the wave equation given by Erwin Schrödinger.