Question

The value of $(n_2+n_1)$ and $(n_2^2-n_1^2)$ for ${He}^{+}$ ion in atomic spectrum are 4 and 8 respectively. The wavelength of emitted photon when electron jump from $n_2$ to $n_1$ is:

• A. $\frac{32}{9}{R}_H$
• B. $\frac{9}{32}{R}_H$
• C. $\frac{9}{32R_H}$
• D. $\frac{32}{9R_H}$

Answer 1

The correct option is C $\frac{9}{32R_H}$

Given that

$n_2+n_1=4$ .....(1)

$n_2^2-n_1^2=8$ -----(2)

Here,

$n_1$ and $n_2$ are the energy levels of $H{e}^{+}$ ion.

Solving equation (2), we have,

$n_2^2-n_1^2=(n_2+n_1)(n_2-n_1)=8$ -----(3)

Substituting (1) in equation (3), we have,

$4\times (n_2-n_1)=8$

$\Rightarrow (n_2-n_1)=\frac{8}{4}=2$ ------- (4)

Solving equations (1) and (4), i.e.

$n_2+n_1=4$ .....(1)

$n_2-n_1=2$ .....(4)

We get,

$n_1=1,n_2=3$

Now, using Rydberg's wavelength equation, we have,

$\frac{1}{\lambda }=R_H{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

where,

$\lambda =$ wavelength of the emitted photon

$R_H=$ Rydberg constant

$Z=$ atomic number of H or H-like ion and

$n_1$ & $n_2$ are the energy levels of H or H-like ion

$\Rightarrow \frac{1}{\lambda }=R_H\times 2^2\times (\frac{1}{1^2}-\frac{1}{3^2})$

$\Rightarrow \frac{1}{\lambda }=R_H\times 2^2\times (\frac{1}{1}-\frac{1}{9})$

$\Rightarrow \frac{1}{\lambda }=R_H\times 2^2\times \left(\frac{9-1}{9}\right)$

$\Rightarrow \frac{1}{\lambda }=R_H\times 4\times \frac{8}{9}=\frac{32R_H}{9}$

$\Rightarrow \lambda =\frac{9}{32R_H}$

$\therefore$ (C) option is correct.