Question

In a sample of H-atoms, electrons de-excite from a level 'n' to 1. The total number of lines belonging to Balmer series are two. If the electrons are ionised from level 'n' by photons of energy 13 eV. Then determine the kinetic energy of the ejected photoelectrons.

• A. 12.15 eV
• B. 13 eV
• C. 15.2 eV
• D. 11.5 eV

Answer 1

The correct option is A 12.15 eV

Balmer series corresponds to transitions upto second energy level. The total number of lines belonging to Balmer series are two.

This corresponds to transitions $4\to 2$ and $3\to 2$.

Hence, $n=4$

The energy for fourth level is $\frac{13.6}{n^2}=\frac{13.6}{4^2}=0.85$.

The kinetic energy of the ejected photoelectrons $=13-0.85=12.15eV$.