Question

When a photon of wavelength $60\text{nm}$ is absorbed by a hydrogen atom, photo-ionization of the atom occurs. The minimum kinetic energy of the emitted electron will be,

• A. $7.1\text{eV}$
• B. $13.6\text{eV}$
• C. $3.6\text{eV}$
• D. $1.9\text{eV}$

Answer 1

The correct option is A $7.1\text{eV}$

Ionization energy of hydrogen atom is $13.6\text{eV}$. The energy of incident photon is,
$E=\frac{hc}{\lambda }=\frac{1243}{60}=20.7\text{eV}$

The least kinetic energy will be, when the electron is in the ground state so that it requires maximum energy to be free.

$\therefore (KE{)}_{min}=20.628-13.6=7.1\text{eV}$

Hence, $\left(A\right)$ is the correct answer.

Key concept : Energy conservation is used here. Incident photon energy is used as ionization energy plus left over energy of $e^{-}.$