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A radiation of wave length $3000 A^{0}$ is required to remove an electron from a metal atom. If a radiation of wave length $2000 A^{0}$ is allowed to incident on the metal surface, the kinetic energy of the emitted electron in kJ/mole is

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$1.98 \times 10^{2}$

$3.3 \times 10^{- 19}$

$3.3 \times 10^{- 20}$

$19.8$

answer is A.

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Detailed Solution

$\frac{h c}{λ} = \frac{h c}{λ_{0}} + K . E$

$K . E = \frac{h c}{λ} - \frac{h c}{λ_{0}}$

$= h c [\frac{1}{λ} - \frac{1}{λ_{0}}]$

$= 6.6 \times 10^{- 34} \times 3 \times 10^{8} (\frac{1}{2000 \times 10^{- 10}} - \frac{1}{3000 \times 10^{- 10}}) = 3.3 x 10^{- 21} = 3.3 x 10^{- 21} x 6.023 x 10^{23}$

$\begin{matrix} K . E = 1.98 \times 10^{2} K J / m o l e \end{matrix}$

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