Question

The ratio of minimum to maximum wavelength of radiation emitted by transition of an electron to ground state of Bohr's hydrogen atom is

• A. $\frac{3}{4}$
• B. $\frac{1}{4}$
• C. $\frac{1}{8}$
• D. $\frac{3}{8}$

Answer 1

Answer: (A)

$\frac{3}{4}$

$\frac{1}{\lambda }=R_{new}(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Where,

Rydberg Constant,$R=\frac{m{e}^4}{8{\epsilon }_o^2{h}^{3}c}$

$\frac{{\lambda }_{max}}{{\lambda }_{min}}=\frac{R(\frac{1}{1^2}-\frac{1}{2^2})}{R(\frac{1}{1^2}-\frac{1}{\infty })}=\frac{3}{4}$

Hence, the ratio of minimum to maximum wavelength of radiation emitted is $\frac{3}{4}$