Question

The ionisation potential of hydrogen atom is $13.6$ volt. In the lowest energy level, this atom is ionised by absorbing a photon of $800\,\mathring{A}$. The kinetic energy of the released electron will be :

• A. 15.51 eV
• B. 2.91 eV
• C. 13.6 eV
• D. 1.91 eV

Answer 1

Answer: (D)

The energy of incident photon in joules is
$=\frac{h c}{\lambda}=\frac{6.62 \times 10^{-34} \times 3 \times 10^{-8}}{800 \times 10^{-10}}$
$=2.4825 \times 10^{-18} J$
The energy of the photon in $eV$ is
$=\frac{2.4825 \times 10^{-18}}{1.6 \times 10^{-19}}=15.51\, eV$
The minimum energy needed to ionise the atom $=13.6\, eV$
Hence, KE of the emitted electron
$=15.51-13.6$
$=1.91\, eV$