If the photon of wavelength 150 pm strikes an atom and one of its...

Total energy $\left(E\right)$ of any photon is given by the relation : $E = \dfrac{hc}{λ}$

Where, $h =$ Planck's constant $= 6 . 6 \times {10}^{- 34}Js$

$c =$ Velocity of light $= 3 \times {10}^{8}m / s$

$\lambda =$ wavelength $= 1 . 5 \times {10}^{- 10}m,$ i.e, \left(150pm\right)$</p><p>Thus,$E = \dfrac{hc}{λ} = \dfrac{6 . 6 \times {10}^{- 34} \times 3 \times {10}^{8}}{1 . 5 \times {10}^{- 10}}; E = 1 . 32 \times {10}^{- 15}J$and, energy of ejected electron \left(E^{'}\right)$

$E^{'} = \dfrac{1}{2}{mv}^{2}$,

where,

$m =$ mass of electron $= 9 . 1 \times {10}^{- 31}kg$

$v =$ velocity of electron

$= 1 . 5 \times {10}^{7}m / s$

$E^{'} = \dfrac{1}{2}{mv}^{2} = \dfrac{1}{2} \times 9 . 1 \times {10}^{- 31} \times \left(1 . 5 \times {10}^{7}\right)2$

$E^{'} = 1 . 024 \times {10}^{- 16}$

Thus, total energy of photon = Binding energy of electron

\left(B\right)$+ energy of ejected electron \left(E^{'}\right)$

Thus, 1 . 32 \times {10}^{- 15} = \left(B\right) + \left( E\right)$</p><p>$∴ \left(B\right) = \left( E\right) - \left(E^{'}\right) = \left[1 . 32 \times {10}^{- 15}\right] - \left[1 . 024 \times {10}^{- 16}\right]

$= 1 . 2176 \times {10}^{- 15}J$

$= \dfrac{1 . 2176 \times {10}^{- 15}}{1 . 6 \times {10}^{- 19}}$ eV = 7 . 6 \times {10}^{3}eV$</p></blockquote>$