An atom absorbs a photon of wavelength 500 nm and emits another photon of wavelength 600 nm. The net energy absorbed by the atom in this process is . The value of is ________. [2023]

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Solution

Q.
An atom absorbs a photon of wavelength 500 nm and emits another photon of wavelength 600 nm. The net energy absorbed by the atom in this process is $n \times 10^{- 4} eV$ . The value of $n$ is ________. [2023]

[Assume the atom to be stationary during the absorption and emission process]

(Take $h = 6.6 \times 10^{- 34} Js$ and $c = 3 \times 10^{8} m / s$ )

Ans.
(4125)

$E = E_{1} - E_{2} = \frac{h c}{λ_{1}} - \frac{h c}{λ_{2}} = h c (\frac{1}{λ_{1}} - \frac{1}{λ_{2}})$

$= 6.6 \times 10^{- 34} \times 3 \times 10^{8} (\frac{1}{500 \times 10^{- 9}} - \frac{1}{600 \times 10^{- 9}})$

$= 6.6 \times 10^{- 20} J$

$= \frac{6.6 \times 10^{- 20}}{1.6 \times 10^{- 19}} eV = 4.125 \times 10^{- 1} eV = 4125 \times 10^{- 4} eV$

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