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Physics MCQs for NEET — Practice Questions with Answers

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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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The order of the size of nucleus and Bohr radius of an atom respectively are

$10^{- 14} m, 10^{- 10} m$

$10^{- 10} m, 10^{- 8} m$

$10^{- 20} m, 10^{- 16} m$

$10^{- 8} m, 10^{- 6} m$

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Explanation

(a) Diameter of nucleus is of the order of $10^{- 14}$ m and radius of first Bohr orbit of hydrogen atom

r = $0.53 \times 10^{- 10} m$

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The ratio of the wavelengths for 2 $\to$ 1 transition in ${Li}^{+ +}$ , ${He}^{+}$ and H is-

1 : 2 : 3

1 : 4 : 9

4 : 9 : 36

3 : 2 : 1

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Explanation

$λ_{{Li}^{+ +}} : λ_{{He}^{+}} : λ_{H} = 4 : 9 : 36$

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The wavelength of light emitted from second orbit to first orbits in a hydrogen atom is

$1.215 \times 10^{- 7} m$

$1.215 \times 10^{- 5} m$

$1.215 \times 10^{- 4} m$

$1.215 \times 10^{- 3} m$

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Explanation

(a) Energy radiated E = 10.2 eV = $10.2 \times 1.6 \times 10^{- 19} J$

$\Rightarrow E = \frac{hc}{λ} \Rightarrow λ = 1.215 \times 10^{- 7} m$

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Energy of the electron in nth orbit of hydrogen atom is given by $E_{n} = - \frac{13.6}{n^{2}} eV$ . The amount of energy needed to transfer electron from first orbit to third orbit is

13.6 eV

3.4 eV

12.09 eV

1.51 eV

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Explanation

(c) For n =1 , $E_{1} = - \frac{13.6}{1^{2}} = - 13.6 eV$
and for n = 3, $E_{3} = \frac{- 13.6}{3^{2}} = - 1.51 eV$
So required energy = $E_{3} - E_{1} = - 1.51 - (- 13.6) = 12.09 eV$

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The de-Broglie wavelength of an electron in the first Bohr orbit is

Equal to one fourth the circumference of the first orbit

Equal to half the circumference of the first orbit

Equal to twice the circumference of the first orbit

Equal to the circumference of the first orbit

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Explanation

(d) ${mvr}_{n} = \frac{nh}{2 π} \Rightarrow {pr}_{n} = \frac{nh}{2 π} \Rightarrow \frac{h}{λ} \times r_{n} = \frac{nh}{2 π}$

$\Rightarrow λ = \frac{2 {πr}_{n}}{n}$ , for first orbit n = 1 so $λ = 2 {πr}_{1}$
= circumference of first orbit

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The frequency of 1st line of Balmer series in $H_{2}$ atom is $v_{0}$ . The frequency of line emitted by singly ionised He atom is

2 $v_{0}$

4 $v_{0}$

$v_{0}$ /2

$v_{0}$ /4

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Explanation

(b) $v \propto Z^{2} \Rightarrow \frac{v_{H_{2}}}{v_{He}} = {(\frac{1}{2})}^{2} = \frac{1}{4} \Rightarrow v_{He} = 4 v_{H_{2}} = 4 v_{0}$

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When the electron in the hydrogen atom jumps from 2nd orbit to 1st orbit, the wavelength of radiation emitted is $λ$ . When the electrons jump from 3rd orbit to 1st orbit, the wavelength of emitted radiation would be

$\frac{27}{32} λ$

$\frac{32}{27} λ$

$\frac{2}{3} λ$

$\frac{3}{2} λ$

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Explanation

(a) $\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

First condition $\frac{1}{λ} = R [\frac{1}{1^{2}} - \frac{1}{2^{2}}] \Rightarrow R = \frac{4}{3 λ}$

Second condition $\frac{1}{λ'} = R [\frac{1}{1^{2}} - \frac{1}{3^{2}}]$

$\Rightarrow λ' = \frac{9}{8 R} \Rightarrow λ' = \frac{9}{8 \times \frac{4}{3 λ}} = \frac{27 λ}{32}$

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Which of the following transition will have shortest emission wavelength ?

n = 2 to n =1

n = 1 to n = 2

n = 2 to n = 5

n = 5 to n = 2

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Explanation

The wavelength of emitted radiation is inversely proportional to the frequency. The transition from n=2 to n=1 in the hydrogen atom has the highest frequency and hence the shortest wavelength emission.

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