The wavelength of a spectral line emitted by hydrogen atom in the Balmer series is $\frac{16}{3 R}$ ( $R$ is Rydberg constant). What is the value of the principal quantum number of the state from which the transition takes place?

• A

  2
• B

  3
• Correct answer
  C

  4
• D

  5

Solution

In spectral line series, wavelength is given by $\frac{1}{\lambda}=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$ For Balmer transitions,
$\Rightarrow \quad \frac{1}{\lambda}=R\left(\frac{1}{4}-\frac{1}{n^2}\right)$ Given, $\quad \lambda=\frac{16}{3 R}$ $\Rightarrow \quad \frac{3 R}{16}=R\left(\frac{1}{4}-\frac{1}{n^2}\right) \Rightarrow \frac{3}{16}=\frac{1}{4}-\frac{1}{n^2}$ $\Rightarrow \quad \frac{1}{n^2}=\frac{1}{4}-\frac{3}{16}=\frac{1}{16} \Rightarrow n=4$

Hence, the correct answer is option: 'C'.