Question

The wavelength of the energy emitted when the electron comes from fourth orbit to second orbit in hydrogen is $20.397\text{cm}$. The wavelength of energy for the same transition in $H{e}^{+}$ is

• A. $5.099\text{cm}$
• B. $20.497\text{cm}$
• C. $40.994\text{cm}$
• D. $81.998\text{cm}$

Answer 1

The correct option is A $5.099\text{cm}$

The formula for wavelength of radiation emitted during a transition from higher energy state $\left(n_i\right)$ to lower energy state $\left(n_f\right)$ is given by,

$\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_f^2}-\frac{1}{n_i^2})$

Here, in this question, the initial and the final states are same in both the cases,

$\Rightarrow \lambda \propto \frac{1}{Z^2}$

$\Rightarrow {\lambda }_1\propto \frac{1}{Z_1^2}$ and ${\lambda }_2\propto \frac{1}{Z_2^2}$

$\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\frac{Z_2^2}{Z_1^2}$

Given that, ${\lambda }_1=20.397\text{cm},Z_1=1,Z_2=2$

$\Rightarrow \frac{20.397\text{cm}}{{\lambda }_2}=\frac{2^2}{1^2}$

$\Rightarrow {\lambda }_2=5.099\text{cm}$

Hence, option $\left(A\right)$ is correct.

Why this Question? This type of questions are often asked in JEE Mains and Advance. This gives a basic idea of energy and wavelength transition from one orbit state to another orbit state.