Question

The energy required for the ionization of a certain atom is $3.44 \times 10^{-18}$ J. The absorption of a photon of unknown wavelength ionizes the atom and ejects an electron with velocity $1 \cdot 0.3 \times 10^{6} ms ^{-1}$. Calculate wavelength of the incident radiation.

Answer 1

SOLUTION : Total encrgy of the incident radiation is used to eject an clectron completely from the atom. This energy is equal to the energy for ionization of the atom and the kinetic energy gained by electron.

Kinctic encrgy of the electron —

$=\frac{1}{2} m v^{2}=\frac{1}{2} \times 9 \cdot 1 \times 10^{-31} \times(1 \cdot 03)^{2} \times 10^{12}=4 \cdot 83 \times 10^{-19} J$$

Total energy, $E=4.83 \times 10^{-19}+3.44 \times 10^{-18}=3.923 \times 10^{-18} J$

Now $\quad \lambda=\frac{h c}{E}=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{3.923 \times 10^{-18}} m =50.67 nm$.