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Atoms MCQs for NEET — Physics Questions with Answers

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In Bohr model of hydrogen atom, the ratio of periods of revolution of an electron in n =2 and n = 1 orbits is

2 : 1

4 : 1

8 : 1

16 : 1

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Explanation

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The ratio of the longest to shortest wavelengths in Brackett series of hydrogen spectra is

$\frac{25}{9}$

$\frac{17}{6}$

$\frac{9}{5}$

$\frac{4}{3}$

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Explanation

(a) For Bracket series $\frac{1}{λ_{\max}} = R [\frac{1}{4^{2}} - \frac{1}{5^{2}}] = \frac{9}{25 \times 16} R$
and $\frac{1}{λ_{\min}} = R [\frac{1}{4^{2}} - \frac{1}{\infty^{2}}] = \frac{R}{16} \Rightarrow \frac{λ_{\max}}{λ_{\min}} = \frac{25}{9}$

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The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statements is true

Its kinetic energy increases and its potential and total energies decrease

Its kinetic energy decreases, potential energy increases and its total energy remains the same

Its kinetic and total energies decrease and its potential energy increases

Its kinetic, potential and total energies decreases

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Explanation

(a) For hydrogen and hydrogen like atoms $E_{n} = - 13.6 \frac{z^{2}}{n^{2}} eV$

$U_{n} = 2 E_{n} = - 27.2 \frac{z^{2}}{n^{2}} eV$ and $K_{n} = |E_{n}| = 13.6 \frac{z^{2}}{n^{2}} eV$

From these three relations we can see that as n decreases, $K_{n}$ will increase but $E_{n}$ and $U_{n}$ will decreases.

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The ratio of minimum to maximum wavelength in Balmer series is

5 : 9

5 : 36

1 : 4

3 : 4

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Explanation

(a) $\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] \Leftarrow \frac{λ_{\min}}{λ_{\max}} = \frac{[\frac{1}{2^{2}} - \frac{1}{3^{2}}]}{[\frac{1}{2^{2}} - \frac{1}{\infty^{2}}]} = \frac{5}{9}$

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Rutherford’s $α$ -particle experiment showed that the atoms have

Proton

Nucleus

Neutron

Electrons

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Explanation

(b) It shows that atoms has a central positively charged nucleus that is surrounded by electrons.

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Which of the following is true for number of spectral lines in going form Layman series to Pfund series

Increases

Decreases

Unchanged

May decreases or increases

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Explanation

(b) Maximum number of spectral lines are observed in Layman series.

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Radius of the first orbit of the electron in a hydrogen atom is 0.53 Å. So, the radius of the third orbit will be

2.12 Å

4.77 Å

1.06 Å

1.59 Å

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Explanation

(b) $r_{n} \propto n^{2} \Rightarrow \frac{r_{3}}{r_{1}} = \frac{3^{2}}{1} \Rightarrow r_{3} = 9 r_{1} = 9 \times 0.53 = 4.77$ Å

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The first line in the Lyman series has wavelength $λ$ . The wavelength of the first line in Balmer series is

$\frac{2}{9} λ$

$\frac{9}{2} λ$

$\frac{5}{27} λ$

$\frac{27}{5} λ$

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Explanation

(d) For first line in Lyman series $λ_{L_{1}} = \frac{4}{3 R}$ ..... (i)
For first line in Balmer series $λ_{B_{1}} = \frac{36}{5 R}$ ..... (ii)
From equation (i) and (ii)

$\frac{λ_{B_{1}}}{λ_{L_{1}}} = \frac{27}{5} \Rightarrow λ_{B_{1}} = \frac{27}{5} λ_{L_{1}} \Rightarrow λ_{B_{1}} = \frac{27}{5} λ$

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In the following transitions, which one has higher frequency

3 – 2

4 – 3

4 – 2

3 – 1

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Explanation

(d) 3 – 1 transition has higher energy so it has higher frequency $(v = \frac{E}{h})$

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An electron jumps from 5th orbit to 4th orbit of hydrogen atom. Taking the Rydberg constant as $10^{7}$ per metre. What will be the frequency of radiation emitted

$6.75 \times 10^{12} Hz$

$6.75 \times 10^{14} Hz$

$6.75 \times 10^{13} Hz$

None of these

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Explanation

$\Rightarrow v = 10^{7} \times (3 \times 10^{8}) [\frac{1}{4^{2}} - \frac{1}{5^{2}}] = 6.75 \times 10^{13} Hz$

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