Question

The wave number for the shortest wavelength transition in the Balmer series of hydrogen atoms is:

• A. $4R_{h}c{m}^{-1}$
• B. $2R_{H}c{m}^{-1}$
• C. $\frac{R_H}{4}c{m}^{-1}$
• D. $\frac{R_H}{2}c{m}^{-1}$

Answer 1

Answer: (C)

$\frac{R_H}{4}c{m}^{-1}$

For shortest wavelength of balmer series electron transition is infinite orbit to 2 orbit

We know

$v^{-}=R_H[1/n_1^2-1/n_2^2]$

$=R_H[1/2^2-1/(d{)}^2]$ where d = infinite

$=R_H/4c{m}^{-1}$