Question

What is the wavelength of light emitted when the electron in hydrogen atom undergoes transition from the energy level with $n=4$ to energy level $n=2$ ?
(given $R_n$ = $109678c{m}^{-1}$ )

• A. 786 nm
• B. 486 nm
• C. 586 nm
• D. 986 nm

Answer 1

The correct option is B 486 nm

According to Balmer formula

Wave number $=R_H[1/n_1^2–1/n_2^2]$

Here $n_1=2,n_2=4,R_H=109678$

Putting these values in the equation we get

Wave number $=109678(1/2^2-1/4^2)=109678\times 3/16$

Also $\lambda =$ 1/ wave number

Therefore $\lambda =16/109678\times 3=486nm$