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The kinetic energy of the electron in an orbit of radius r in hydrogen atom is (e = electronic charge)

$\frac{e^{2}}{r^{2}}$

$\frac{e^{2}}{2 r}$

$\frac{e^{2}}{r}$

$\frac{e^{2}}{2 r^{2}}$

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Explanation

(b) Potential energy of electron in nth orbit of radius r in H-atom $U = - \frac{e^{2}}{r}$ (in CGS)

$∵ K . E . = \frac{1}{2} |P . E .| \Rightarrow K = \frac{e^{2}}{2 r}$

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The angular momentum of electron in nth orbit is given by

$\frac{h}{2 πn}$

$n \frac{h}{2 π}$

$n^{2} \frac{h}{2 π}$

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Explanation

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The ratio of the energies of the hydrogen atom in its first to second excited state is

1/ 4

4/9

9/ 4

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Explanation

$∵ E = - \frac{13.6}{n^{2}} \Rightarrow \frac{E_{2}}{E_{3}} = {(\frac{3}{2})}^{2} = \frac{9}{4}$

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An electron jumps from the 4th orbit to the 2nd orbit of hydrogen atom. Given the Rydberg's constant R = $10^{5} {cm}^{- 1}$ . The frequency in Hz of the emitted radiation will be

$\frac{3}{16} \times 10^{5}$

$\frac{3}{16} \times 10^{15}$

$\frac{9}{16} \times 10^{15}$

$\frac{3}{4} \times 10^{15}$

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Explanation

(c) $\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{4^{2}}) = \frac{3 R}{16} \Rightarrow λ = \frac{16}{3 R} = \frac{16}{3} \times 10^{- 5} cm$

Frequency $n = \frac{c}{λ} = \frac{3 \times 10^{10}}{\frac{16}{3} \times 10^{- 5}} = \frac{9}{16} \times 10^{15} Hz$

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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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