Question

The Balmer series in the hydrogen spectrum corresponds to the transition from $n_1=2$ to $n_2=3$, $4$, $........$. This series lies in the visible region. What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with $n=4$ to an energy level with $n=2$?

• A. $486nm$
• B. $386nm$
• C. $846nm$
• D. $648nm$

Answer 1

Answer: (A)

According to Rydberg equation,
$\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
Here, $n_1=2,n_2=4$ and $R=109677c{m}^{-1}$
$\therefore \frac{1}{\lambda }=109677\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=109677\left(\frac{1}{4}-\frac{1}{16}\right)$
$=109677\left(\frac{4-1}{16}\right)$
$\lambda =\frac{16}{109677\times 3}=4.86\times 10^{-5}cm$
$=486\times 10^{-9}m=486nm$
$\left[\because 1nm=10^{-9}m\right]$