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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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In terms of Rydberg's constant R, the wave number of the first Balmer line is

$\frac{5 R}{36}$

$\frac{8 R}{9}$

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Explanation

(c) Wave number = $\frac{1}{λ} = R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
For first Balmer line $n_{1}$ = 2, $n_{2}$ =3
Wave number

$∴$ Wave number = $R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = R (\frac{9 - 4}{9 \times 4}) = \frac{5 R}{36}$

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Which of the transitions in hydrogen atom emits a photon of lowest frequency (n = quantum number)

n = 2 to n = 1

n = 4 to n = 3

n = 3 to n = 1

n = 4 to n = 2

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Explanation

The frequency of the emitted photon is inversely proportional to the square of the principal quantum number (n). The transition from n=4 to n=3 involves the smallest energy difference and hence the lowest frequency photon emission.

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According to Bohr's theory, the expressions for the kinetic and potential energy of an electron revolving in an orbit is given respectively by

$+ \frac{e^{2}}{8 {πε}_{0} r}$ and $- \frac{e^{2}}{4 {πε}_{0} r}$

$+ \frac{8 {πε}_{0} e^{2}}{r}$ and $- \frac{4 {πε}_{0} e^{2}}{r}$

$- \frac{e^{2}}{8 {πε}_{0} r}$ and $- \frac{e^{2}}{4 {πε}_{0} r}$

$+ \frac{e^{2}}{8 {πε}_{0} r}$ and $+ \frac{e^{2}}{4 {πε}_{0} r}$

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Explanation

(a) $P . E . = - \frac{{ke}^{2}}{r} = - \frac{e^{2}}{4 {πε}_{0} r}; K . E = - \frac{1}{2} (PE) = \frac{e^{2}}{8 {πε}_{0} r}$

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Ratio of the wavelengths of first line of Lyman series and first line of Balmer series is

1: 3

27 : 5

5 : 27

4 : 9

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Explanation

(c) $\frac{1}{λ} = R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
For first line of Lymen series $n_{1}$ = 1 and $n_{2}$ = 2
For first line of Balmer series $n_{1}$ = 2 and $n_{2}$ = 3
So, $\frac{λ_{Lyman}}{λ_{Balmer}} = \frac{5}{27}$

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