Question

The spectral line of the shortest wavelength in Balmer series of atomic hydrogen will be:

• A. $6.215\times {10}^{-5}\text{cm}$
• B. $1.437\times {10}^{-5}\text{cm}$
• C. $5.942\times {10}^{-5}\text{cm}$
• D. $3.647\times {10}^{-5}\text{cm}$

Answer 1

The correct option is D $3.647\times {10}^{-5}\text{cm}$

Rydberg formula for hydrogen atom is given by
$\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
For Balmer series $n_1=2$ and for minimum wavelength $n_2=\infty$
Thus the shortest wavelength will be
$\frac{1}{{\lambda }_{shortest}}=109678\times 1^2\times (\frac{1}{2^2}-\frac{1}{{\infty }^2}){\lambda }_{shortest}=3.647\times {10}^{-5}\text{cm}$