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. $486 \,nm$
• B. $386 \,nm$
• C. $846 \,nm$
• D. $648 \,nm$

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 $R = 109677\, cm^{-1}$
$\therefore \, \frac{1}{\lambda}=109677\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=109677 \left(\frac{1}{4}-\frac{1}{16}\right)$
$=109677 \left(\frac{4-1}{16}\right)$
$\lambda=\frac{16}{109677\times3}=4.86\times10^{-5}\, cm$
$=486\times10^{-9}\,m=486\,nm$
$\left[\because\, 1\,nm=10^{-9}\,m\right]$