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An electron has a mass of $9.1 \times 10^{- 31} kg$ . It revolves round the nucleus in a circular orbit of radius $0.529 \times 10^{- 10}$ metre at a speed of $2.2 \times 10^{6} m / s$ . The magnitude of its linear momentum in this motion is

$1.1 \times 10^{- 34}$ kg-m/s

$2.0 \times 10^{- 24}$ kg-m/s

$4.0 \times 10^{- 24}$ kg-m/s

$4.0 \times 10^{- 31}$ kg-m/s

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Explanation

(b) Linear momentum = mv = $9.1 \times 10^{- 31} \times 2.2 \times 10^{6}$

= $2.0 \times 10^{- 24}$ kg-m/s

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The ionization potential for second He electron is

13.6 eV

27.2 eV

54.4 eV

100 eV

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Explanation

(c) For the ionization of second He electron. ${He}^{+}$ will act as hydrogen like atom.
Hence ionization potential $= Z^{2} \times 13.6 volt = {(2)}^{2} \times 13.6 = 54.4 V$

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The energy required to remove an electron in a hydrogen atom from n = 10 state is

13.6 eV

1.36 eV

0.136 eV

0.0136 eV

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Explanation

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Every series of hydrogen spectrum has an upper and lower limit in wavelength. The spectral series which has an upper limit of wavelength equal to 18752 Å is

Balmer series

Lyman series

Paschen series

Pfund series

(Rydberg constant R = $1.097 \times 10^{7}$ per metre)

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Explanation

(c) $\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] \Rightarrow \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} = \frac{1}{Rλ} = \frac{1}{1.097 \times 10^{7} \times 18752 \times 10^{- 10}} = 0.0486 = \frac{7}{144}$

But $\frac{1}{3^{2}} - \frac{1}{4^{2}} = \frac{7}{144} \Rightarrow n_{1} = 3 and n_{2} = 4$ (Paschen series)

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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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