Structure of Atoms

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Ans: The electronic configuration of nickel is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s² , thus it loses electrons from its 4s orbital as its energy is higher compared to the other orbital.

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Ans: The electronic configuration of oxygen atom is 1s² 2s2 2p⁴ and its orbital diagram is given by

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Ans: The effective nuclear charge (Z_eff) is defined as the net positive charge experienced by the outermost electrons in the atom. With the increase of Azimuthal quantum number (l) the Z_eff experienced by the electron decreases. Hence the arrangement of subshells in the increasing order of Z_eff is : d

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Ans: The drawbacks of Bohr's model were

(i) It was unable to explain the spectra for multi-electron systems

(ii) It could not explain the molecule formation through chemical bonds. The two important developments that contributed significantly towards the change of concept of movement of an electron in an orbit was replaced by, the concept of probability of finding an electron in an orbital were

(i) Dual nature of matter

(ii) Uncertainty Principle. Quantum mechanical model of the atom is the name of the new model.

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Ans: We have

v= 109677 [1/n_i ² - 1/n_f ² ]

Given, n_i = 3 and n_f =2

D E= hcv = 109677 [1/n_i ² - 1/n_f ² ]

DE = - 3.052 * 10^-19J

n =DE /h= 4.606 * 10¹⁶Hz

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Ans: The points of the Bohr's model that can be consider are as follows: -

(i) Electrons revolve around the nucleus in a fixed orbit with fixed energy

(ii) The energy is absorbed or released when the electron moves from one energy level to another. The energy for the n^th stationary state is given by

En = -2 p ²me⁴ /n²h²

Where

 m = mass of the electron

e = charge of the electron

 h= Planck's constant If an electron jumps from n_i to n_f then we have

D E= E_f - E_i = 2p²me⁴ /h² [ (1/n_i ² ) - (1/n_f² )]

v = DE /hc = 109677 [ (1/n_i ² ) - (1/n_f² )]

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ANS- For the emission of electrons from metal the frequency of the striking light should be higher than that of its threshold frequency. We have

h _ν = hν ₀ + K.E

ν ₀ = ν - K.E/h ……. (1)

Given

 ν = 10¹⁵ s ^-1, K.E = 1.988 X 10^-19J

Thus (1) gives

ν ₀ = 7 x 10¹⁴ s ^-1

Given

λ = 600 nm

ν = c/ λ = 5 x 10¹⁴ s ^-1

 As ν ₀ > ν, thus the electrons do not emit.