The wavelength of the first spectral line in the Balmer series of the hydrogen atom is $6561 \stackrel{\circ }{\mathrm{A}}$. The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is

• Correct answer
  A

  $1215 \stackrel{\circ }{\mathrm{A}}$
• B

  $1640 \stackrel{\circ }{\mathrm{A}}$
• C

  $2430 \stackrel{\circ }{\mathrm{A}}$
• D

  $4687 \stackrel{\circ }{\mathrm{A}}$

Solution

           

For hydrogen or hydrogen type atoms

           $\frac{1}{\lambda }={RZ}^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$

In the transition from $n_i\to n_f$

$\therefore \lambda \propto \frac{1}{Z^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)}$

$\therefore \frac{{\lambda }_2}{{\lambda }_1}= \frac{Z_1^2{\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)}_1}{Z_2^2{\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)}_2}$

${\lambda }_2= \frac{{\lambda }_1{Z}_1^2{\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)}_1}{Z_2^2{\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)}_2}$

Substituting the values, we have

$=\frac{\left(6561\right){\left(1\right)}^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right)}{{\left(2\right)}^2\left(\frac{1}{2^2} - \frac{1}{4^2}\right)}$ $=1215 \stackrel{\circ }{\mathrm{A}}$

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