Question

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, calculate the energy with which it is bound to the nucleus.

Answer 1

The energy of the incident photon (E) is given by,
E =$\frac{hc}{\lambda}$
= $\frac{(6.626 × 10 ^{- 34} Js)(3.0 × 10^8 ms^{- 1})}{(150 × 10^{-12} m)}$
= 1.3252 × 10-15 J
≈ 1.3252 × 10-16 J
The energy of the electron ejected (K.E) = $\frac{1}{2}$mev2 = $\frac{1}{2}$ (9.10939 × 10-31kg)(1.5 × 107 ms-1)2
= 10.2480 × 1017 J = 1.025 × 1016 J
Hence, the energy with which the electron is bound to the nucleus can be obtained as E = K.E
= 13.252 × 10 -16J -1.025 × 1016 J
= 12.227 × 1016 J
= $\frac{12.227 × 10^{-16}}{1.602 × 10^{-19}}$ eV = 7.6 × 103 eV
$\frac{5\lambda_0-2000}{4\lambda_0-2000}$ = $(\frac{5.35}{2.55})^2$ = $\frac{28.6225}{6.5025}$
$\frac{5\lambda_0-2000}{4\lambda_0-2000}$ = 4.40177
17.6070λ0 - 5λ0 = 8803.537 - 2000
λ0 = $\frac{6805.537}{12.607}$
λ0 = 539.8 nm
λ0 ≈ 540 nm