Question

Find the value of wave number $\overline{v}$ in terms of Rydberg's constant, when transition of electron takes place between two levels of ${He}^{+}$ ion whose sum is $4$ and difference is $2$.

• A. $\frac{8R}{9}$
• B. $\frac{32R}{9}$
• C. $\frac{3R}{4}$
• D. None of these

Answer 1

The correct option is B $\frac{32R}{9}$

Rydberg's formula is $\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_f^2}-\frac{1}{n_i^2})$ ${}_V^{-}$ = $\frac{1}{\lambda }$,

We have given that $n_f-n_i=2$ And $n_f+n_i=4$

Solving these equation $n_f=3$ $n_i=1$

For helium, $Z=2$-

Using rydberg's formula,

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

${}_V^{-}$ $=\frac{32R}{9}$

Therefore, the answer is B.