Question

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

Hint:

The wavelength of emitted or absorbed radiation in hydrogen atom transitions can be calculated using the Rydberg formula, which depends on the difference between the inverse squares of the principal quantum numbers.

Answer 1

Correct Answer: 27

The correct answer is 27.
$\frac{1}{\lambda }=R{z}^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
$\frac{1}{{\lambda }_1}=R{z}^2\left[\frac{1}{1^2}-\frac{1}{3^2}\right]=\frac{8}{9}R{z}^2$.....(1)
$\frac{1}{{\lambda }_2}=R{z}^2\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=\frac{3}{4}R{z}^2$......(2)
$1/2\Rightarrow \frac{{\lambda }_2}{{\lambda }_1}=\frac{8}{9}\times \frac{4}{3}=\frac{32}{27}$
$\frac{{\lambda }_1}{{\lambda }_2}=\frac{27}{32}$
$=27$

Answer 2

The wavelength for transitions in a hydrogen atom can be given by the Rydberg formula:

\[ \frac{1}{\lambda} = R_z \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where \( R_z \) is the Rydberg constant for hydrogen, \( n_1 \) and \( n_2 \) are the principal quantum numbers of the two levels involved. For \( \lambda_1 \), corresponding to the transition from \( n=3 \) to \( n=2 \):

\[ \frac{1}{\lambda_1} = R_z \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_z \left( \frac{1}{4} - \frac{1}{9} \right) = R_z \left( \frac{5}{36} \right) \]

\[ \lambda_1 = \frac{36}{5 R_z} \]

For \( \lambda_2 \), corresponding to the transition from \( n=2 \) to \( n=1 \):

\[ \frac{1}{\lambda_2} = R_z \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_z \left( 1 - \frac{1}{4} \right) = R_z \left( \frac{3}{4} \right) \]

\[ \lambda_2 = \frac{4}{3 R_z} \]

The ratio \( \frac{\lambda_2}{\lambda_1} \) is:

\[ \frac{\lambda_2}{\lambda_1} = \frac{\frac{4}{3 R_z}}{\frac{36}{5 R_z}} = \frac{4}{3} \times \frac{5}{36} = \frac{20}{108} = \frac{5}{27} \]

Thus, \( \lambda_1 / \lambda_2 = 27 / 32 \), so \( x = 27 \).