For hydrogen atom, $$\lambda_1$$ and $$\lambda_2$$ are the wavelengths corresponding to the transitions 1 and 2 respectively as shown in figure. The ratio of $$\lambda_1$$ and $$\lambda_2$$ is $$\dfrac{x}{32}$$. The value of $$x$$ is ______.

Solution :

For hydrogen atom,

$$\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

Transition 1 :

$$n_2 = 3,\ n_1 = 1$$

Therefore,

$$\frac{1}{\lambda_1} = R\left(1 - \frac{1}{9}\right)$$

$$= R \times \frac{8}{9}$$

Transition 2 :

$$n_2 = 2,\ n_1 = 1$$

Therefore,

$$\frac{1}{\lambda_2} = R\left(1 - \frac{1}{4}\right)$$

$$= R \times \frac{3}{4}$$

Now,

$$\frac{\lambda_1}{\lambda_2}=\frac{\frac{1}{R(8/9)}}{\frac{1}{R(3/4)}}$$

$$=\frac{3/4}{8/9}$$

$$=\frac{3}{4}\times\frac{9}{8}$$

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

Given,

$$\frac{\lambda_1}{\lambda_2} = \frac{x}{32}$$

Therefore,

$$x = 27$$

Final Answer :

$$27$$