Question

In a hydrogen atom, the transition takes place from $n=3$ to $n=2$. If Rydberg constant is $1.097\times {10}^7{m}^{-1}$, the wavelength of the emitted radiation is:

• A. $6564\mathring{A}$
• B. $6064\mathring{A}$
• C. $6664\mathring{A}$
• D. $5664\mathring{A}$

Answer 1

The correct option is A $6564\mathring{A}$

$\frac{1}{\lambda }=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$=1.097\times {10}^7[\frac{1}{2^2}-\frac{1}{3^2}]=0.152\times {10}^7$

$\therefore \lambda =\frac{1}{0.152\times {10}^7}=6.564\times {10}^{-7}m=6564\mathring{A}$