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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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If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of ${Li}^{+ +}$ is

122.4 eV

30.6 eV

13.6 eV

3.4 eV

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Explanation

(b) $E_{n} = \frac{13.6}{n^{2}} \times Z^{2}$

For first excited state n = 2 and for ${Li}^{+ +}$ , z =3 $\Rightarrow E = \frac{13.6}{4} \times 9 = 30.6 eV$

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Which state of triply ionised Beryllium $({Be}^{+ + +})$ has the same orbital radius as that of the ground state of hydrogen

n = 4

n = 3

n = 2

n = 1

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Explanation

(c) Radius of nth orbit for any hydrogen like atom
$r_{n} = r_{0} (\frac{n^{2}}{Z})$ ( $r_{0}$ = radius of first orbit of $H_{2}$ -atom)
If $r_{n} = r_{0} \Rightarrow n = \sqrt{Z}$ . For ${Be}^{+ + +}$ , Z = 4 $\Rightarrow$ n = 2

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Taking Rydberg’s constant $R_{H} = 1.097 \times 10^{7} m$ first and second wavelength of Balmer series in hydrogen spectrum is

2000 Å, 3000 Å

1575 Å, 2960 Å

6529 Å, 4280 Å

6552 Å, 4863 Å

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Explanation

(d) $\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$ For first wavelength, $n_{1} = 2$ , $n_{2} = 3$ $\Rightarrow λ_{1} = 6563$ Å. For second wavelength, $n_{1} = 2, n_{2} = 4 \Rightarrow λ_{2} = 4861$ Å

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The kinetic energy of electron in the first Bohr orbit of the hydrogen atom is

– 6.5 eV

– 27.2 eV

13.6 eV

– 13.6 eV

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Explanation

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In Bohr’s model of hydrogen atom, which of the following pairs of quantities are quantized

Energy and linear momentum

Linear and angular momentum

Energy and angular momentum

None of the above

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Explanation

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