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Atoms MCQs for NEET — Physics Questions with Answers

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According to Bohr's theory the radius of electron in an orbit described by principal quantum number n and atomic number Z is proportional to

$Z^{2} n^{2}$

$\frac{Z^{2}}{n^{2}}$

$\frac{Z^{2}}{n}$

$\frac{n^{2}}{Z}$

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Explanation

(d) $r = \frac{ε_{0} n^{2} h^{2}}{{πZme}^{2}}; ∴ r \propto \frac{n^{2}}{Z}$

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The radius of electron's second stationary orbit in Bohr's atom is R. The radius of the third orbit will be

3 R

2.25 R

9 R

$\frac{R}{3}$

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Explanation

(b) $r \propto n^{2} \Rightarrow \frac{n_{(n = 2)}}{n_{(n = 3)}} = \frac{4}{9} \Rightarrow r_{(n = 3)} = \frac{9}{4} R = 2.25 R$

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In any Bohr orbit of the hydrogen atom, the ratio of kinetic energy to potential energy of the electron is

1/2

$- 1 / 2$

– 2

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Explanation

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The spectral series of the hydrogen spectrum that lies in the ultraviolet region is the

Balmer series

Pfund series

Paschen series

Lyman series

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Explanation

(d) Lyman series lies in the UV region.

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A hydrogen atom (ionisation potential 13.6 eV) makes a transition from third excited state to first excited state. The energy of the photon emitted in the process is

1.89 eV

2.55 eV

12.09 eV

12.75 eV

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Explanation

(b) Energy released = $13.6 [\frac{1}{{(2)}^{2}} - \frac{1}{{(4)}^{2}}] = 2.55 eV$

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When a hydrogen atom is raised from the ground state to an excited state

P.E. increases and K.E. decreases

P.E. decreases and K.E. increases

Both kinetic energy and potential energy increase

Both K.E. and P.E. decrease

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Explanation

(a) $P . E \propto - \frac{1}{r} and K . E . \propto \frac{1}{r}$

As r increases so K.E. decreases but P.E. increases.

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The ratio of the kinetic energy to the total energy of an electron in a Bohr orbit is

– 1

1 : 2

None of these

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Explanation

(a) K.E. = – (T.E.)

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An electron in the n = 1 orbit of hydrogen atom is bound by 13.6 eV. If a hydrogen atom is in the n = 3 state, how much energy is required to ionize it

13.6 eV

4.53 eV

3.4 eV

1.51 eV

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Explanation

(d) Required energy $E_{3} = + \frac{13.6}{3^{2}} = 1.51 eV$

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The ratio of the frequencies of the long wavelength limits of Lyman and Balmer series of hydrogen spectrum is

27 : 5

5 : 27

4 : 1

1 : 4

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Explanation

(a) For Lyman series

$v_{Lyman} = \frac{c}{λ_{\max}} = Rc [\frac{1}{{(1)}^{2}} - \frac{1}{{(2)}^{2}}] = \frac{3 RC}{4}$

For Balmer series

$v_{B a l m e r} = \frac{c}{λ_{\max}} = Rc [\frac{1}{{(2)}^{2}} - \frac{1}{{(3)}^{2}}] = \frac{5 RC}{36}$

$∴ \frac{v_{Lyman}}{v_{Balmer}} = \frac{27}{5}$

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Which of the following transitions in a hydrogen atom emits photon of the highest frequency

n = 1 to n = 2

n = 2 to n = 1

n = 2 to n = 6

n = 6 to n = 2

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Explanation

The frequency of a photon emitted during a transition is inversely proportional to the wavelength. The transition from n=2 to n=1 in the hydrogen atom corresponds to the shortest wavelength in the Balmer series, which means it has the highest frequency.

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