If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 × 107 ms–1,

Solution:

Energy of incident photon $(E)$ is given by,

$E=\frac{h c}{\lambda}$

$=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.0 \times 10^{8} \mathrm{~ms}^{-1}\right)}{\left(150 \times 10^{-12} \mathrm{~m}\right)}$

$=1.3252 \times 10^{-15} \mathrm{~J}$

$\simeq 13.252 \times 10^{-16} \mathrm{~J}$

Energy of the electron ejected (K.E)

$=\frac{1}{2} \mathrm{~m}_{e} v^{2}$

$=\frac{1}{2}\left(9.10939 \times 10^{-31} \mathrm{~kg}\right)\left(1.5 \times 10^{7} \mathrm{~ms}^{-1}\right)^{2}$

$=10.2480 \times 10^{-17} \mathrm{~J}$

$=1.025 \times 10^{-16} \mathrm{~J}$

Hence, the energy with which the electron is bound to the nucleus can be obtained as:

$=E-K \cdot E$

$=13.252 \times 10^{-16} \mathrm{~J}-1.025 \times 10^{-16} \mathrm{~J}$

$=12.227 \times 10^{-16} \mathrm{~J}$

$=\frac{12.227 \times 10^{-16}}{1.602 \times 10^{-19}} \mathrm{eV}$

$=7.6 \times 10^{3} \mathrm{eV}$

$\frac{5 \lambda_{0}-2000}{4 \lambda_{0}-2000}=\left(\frac{5.35}{2.55}\right)^{2}=\frac{28.6225}{6.5025}$

$\frac{5 \lambda_{0}-2000}{4 \lambda_{0}-2000}=4.40177$

$17.6070 \lambda_{0}-5 \lambda_{0}=8803.537-2000$

$\lambda_{0}=\frac{6805.537}{12.607}$

$\lambda_{0}=539.8 \mathrm{~nm}$

$\lambda_{0}=540 \mathrm{~nm}$