### Atomic Physics Question 6

#### Question 6 - 31 January - Shift 1

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 $\frac{x}{32}$. The value of $x$ is_______

## Show Answer

#### Answer: (27)

#### Solution:

#### Formula: Wavelength Corresponding To Spectral Lines

$ \begin{aligned} & \begin{matrix} 1 \\ \lambda \end{matrix} =Rz^{2} \begin{bmatrix} 1 & 1 \\ n_1^{2} & -n_2^{2} \end{bmatrix} \\ & 1 \\ & \lambda_1=Rz^{2} \begin{bmatrix} 1 & 1 \\ 1^{2} & 3^{2} \end{bmatrix} = _9^{8} Rz^{2} \\ & 1=Rz^{2} \begin{bmatrix} 1 & 1 \\ 1^{2} & 2^{2} \end{bmatrix} = _4^{3} Rz^{2} \\ & \frac{1}{2} \Rightarrow \frac{\lambda_2}{\lambda_1}= _9^{8} \times _3^{4}= _27^{32} \\ & \frac{\lambda_1}{\lambda_2}=\frac{27}{32} \end{aligned} $

Ans. 27