An atom absorbs a photon of wavelength 500nm and emits another photon of wavelength 600nm. The net energy absorbed by the atom in this process is n x 10^-4eV .The value of n is _______
(h = 6.6 x 10^-34 Js and c = 3 x 10⁸m/s)

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answer is 4125.

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Detailed Solution

Energy absorved $= \frac{hc}{λ_{1}} - \frac{hc}{λ_{2}}$
$\begin{array}{l} = \frac{hc}{λ_{1}} - \frac{hc}{λ_{2}} \\ = hc (\frac{1}{λ_{1}} - \frac{1}{λ_{2}}) \\ = 6.6 \times 10^{- 31} \times 3 \times 10^{8} (\frac{1}{500 \times 10^{- 9}} - \frac{1}{600 \times 10^{- 9}}) J \\ = \frac{6.6 \times 3 \times 10^{- 19}}{1.6 \times 10^{- 19}} (\frac{1}{5} - \frac{1}{6}) ev \\ = \frac{6.6 \times 3}{1.6 \times 30} ev \\ = 0.4125 ev \\ = 4125 \times 10^{- 4} ev \end{array}$