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The ratio of minimum to maximum wavelength in Balmer series is

5 : 9

5 : 36

1 : 4

3 : 4

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Explanation

(a) $\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] \Leftarrow \frac{λ_{\min}}{λ_{\max}} = \frac{[\frac{1}{2^{2}} - \frac{1}{3^{2}}]}{[\frac{1}{2^{2}} - \frac{1}{\infty^{2}}]} = \frac{5}{9}$

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Rutherford’s $α$ -particle experiment showed that the atoms have

Proton

Nucleus

Neutron

Electrons

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Explanation

(b) It shows that atoms has a central positively charged nucleus that is surrounded by electrons.

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Which of the following is true for number of spectral lines in going form Layman series to Pfund series

Increases

Decreases

Unchanged

May decreases or increases

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Explanation

(b) Maximum number of spectral lines are observed in Layman series.

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Radius of the first orbit of the electron in a hydrogen atom is 0.53 Å. So, the radius of the third orbit will be

2.12 Å

4.77 Å

1.06 Å

1.59 Å

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Explanation

(b) $r_{n} \propto n^{2} \Rightarrow \frac{r_{3}}{r_{1}} = \frac{3^{2}}{1} \Rightarrow r_{3} = 9 r_{1} = 9 \times 0.53 = 4.77$ Å

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The first line in the Lyman series has wavelength $λ$ . The wavelength of the first line in Balmer series is

$\frac{2}{9} λ$

$\frac{9}{2} λ$

$\frac{5}{27} λ$

$\frac{27}{5} λ$

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Explanation

(d) For first line in Lyman series $λ_{L_{1}} = \frac{4}{3 R}$ ..... (i)
For first line in Balmer series $λ_{B_{1}} = \frac{36}{5 R}$ ..... (ii)
From equation (i) and (ii)

$\frac{λ_{B_{1}}}{λ_{L_{1}}} = \frac{27}{5} \Rightarrow λ_{B_{1}} = \frac{27}{5} λ_{L_{1}} \Rightarrow λ_{B_{1}} = \frac{27}{5} λ$

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In the following transitions, which one has higher frequency

3 – 2

4 – 3

4 – 2

3 – 1

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Explanation

(d) 3 – 1 transition has higher energy so it has higher frequency $(v = \frac{E}{h})$

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An electron jumps from 5th orbit to 4th orbit of hydrogen atom. Taking the Rydberg constant as $10^{7}$ per metre. What will be the frequency of radiation emitted

$6.75 \times 10^{12} Hz$

$6.75 \times 10^{14} Hz$

$6.75 \times 10^{13} Hz$

None of these

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Explanation

$\Rightarrow v = 10^{7} \times (3 \times 10^{8}) [\frac{1}{4^{2}} - \frac{1}{5^{2}}] = 6.75 \times 10^{13} Hz$

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The order of the size of nucleus and Bohr radius of an atom respectively are

$10^{- 14} m, 10^{- 10} m$

$10^{- 10} m, 10^{- 8} m$

$10^{- 20} m, 10^{- 16} m$

$10^{- 8} m, 10^{- 6} m$

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Explanation

(a) Diameter of nucleus is of the order of $10^{- 14}$ m and radius of first Bohr orbit of hydrogen atom

r = $0.53 \times 10^{- 10} m$

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The ratio of the wavelengths for 2 $\to$ 1 transition in ${Li}^{+ +}$ , ${He}^{+}$ and H is-

1 : 2 : 3

1 : 4 : 9

4 : 9 : 36

3 : 2 : 1

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Explanation

$λ_{{Li}^{+ +}} : λ_{{He}^{+}} : λ_{H} = 4 : 9 : 36$

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The wavelength of light emitted from second orbit to first orbits in a hydrogen atom is

$1.215 \times 10^{- 7} m$

$1.215 \times 10^{- 5} m$

$1.215 \times 10^{- 4} m$

$1.215 \times 10^{- 3} m$

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Explanation

(a) Energy radiated E = 10.2 eV = $10.2 \times 1.6 \times 10^{- 19} J$

$\Rightarrow E = \frac{hc}{λ} \Rightarrow λ = 1.215 \times 10^{- 7} m$

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