For hydrogen atom, ' λ 1 ' and ' λ 2 ' are the wavelengths corresponding to the transitions 1 and 2 respectively as shown in figure. The ratio of ' λ 1 ' and '

Question

For hydrogen atom, ' λ 1 ​ ' and ' λ 2 ​ ' are the wavelengths corresponding to the transitions 1 and 2 respectively as shown in figure. The ratio of ' λ 1 ​ ' and ' λ 2 ​ ' is 32 x ​ . The value of ' x ' is

TestHub · Accepted Answer

Using Rydberg's formula for the wavelength of an emitted photon:

$\frac{1}{λ} = R (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}})$

For transition 1 ( $n = 3 \to n = 1$ ):

$\frac{1}{λ _{1}} = R (\frac{1}{1 ^{2}} - \frac{1}{3 ^{2}}) = R (1 - \frac{1}{9}) = \frac{8 R}{9}$

$\Rightarrow λ_{1} = \frac{9}{8 R}$

For transition 2 ( $n = 2 \to n = 1$ ):

$\frac{1}{λ _{2}} = R (\frac{1}{1 ^{2}} - \frac{1}{2 ^{2}}) = R (1 - \frac{1}{4}) = \frac{3 R}{4}$

$\Rightarrow λ_{2} = \frac{4}{3 R}$

Taking the ratio of $λ_{1}$ and $λ_{2}$ :

$\frac{λ _{1}}{λ _{2}} = \frac{\frac{9}{8 R}}{\frac{4}{3 R}} = \frac{9}{8} \times \frac{3}{4} = \frac{27}{32}$

Given that $\frac{λ _{1}}{λ _{2}} = \frac{x}{32}$ , we get:

$\frac{x}{32} = \frac{27}{32}$

$\Rightarrow x = 27$

Answer: 27

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