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The ionisation potential of hydrogen atom is 13.6 volt. The energy required to remove an electron in the n = 2 state of the hydrogen atom is

27.2 eV

13.6 eV

6.8 eV

3.4 eV

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Explanation

(d) Energy required to remove electron in the n = 2 state = $+ \frac{13.6}{{(2)}^{2}} = + 3.4 eV$

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If the wavelength of the first line of the Balmer series of hydrogen is 6561 $\overset{0}{A}$ , the wavelength of the second line of the series should be

13122 $\overset{0}{A}$

3280 $\overset{0}{A}$

4860 $\overset{0}{A}$

2187 $\overset{0}{A}$

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Explanation

(c) The wavelength of spectral line in Balmer series is given by $\frac{1}{λ} = R [\frac{1}{2^{2}} - \frac{1}{n^{2}}]$
For first line of Balmer series, n = 3

$\Rightarrow \frac{1}{λ_{1}} = R [\frac{1}{2^{2}} - \frac{1}{3^{2}}] = \frac{5 R}{36}$ ; For second line n = 4.

$\Rightarrow \frac{1}{λ_{2}} = R [\frac{1}{2^{2}} - \frac{1}{4^{2}}] = \frac{3 R}{36}$

$∴ \frac{λ_{2}}{λ_{1}} = \frac{20}{27} \Rightarrow λ_{1} = \frac{20}{27} \times 6561 = 4860 \overset{0}{A}$

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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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