Question

Calculate the wavenumber of the spectral line emitted when the electron de-excites from n = 3 to the ground state in the hydrogen spectrum?
$(R_H=109678c{m}^{-1}$)

• A. $79491.56c{m}^{-1}$
• B. $97491.56c{m}^{-1}$
• C. $91471.56c{m}^{-1}$
• D. $46491.56c{m}^{-1}$

Answer 1

The correct option is B $97491.56c{m}^{-1}$

Using the formula,
$\overline{\nu }=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})\text{where}\lambda \text{is the wavelength of the radiation.}$
$n_1$ = 1 (ground state)
$n_2$ = 3 (excited state)
$R_H=109678c{m}^{-1};n_1=1;n_2=3$
$\overline{\nu }=109678[\frac{1}{1^2}-\frac{1}{3^2}]=109678\times \frac{8}{9}c{m}^{-1}$
On solving, $\overline{\nu }=97491.56c{m}^{-1}$