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 $\mathrm{n} \times 10^{-4} \mathrm{eV}$. The value of n is ______. [Assume the atom to be stationary during the absorption and emission process] (Take $\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}$ and $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ ).

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Modern Physics

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 1 0^{- 4} eV

. The value of n is ______.
[Assume the atom to be stationary during the absorption and emission process]
(Take

h = 6.6 \times 1 0^{- 34} Js

and

c = 3 \times 1 0^{8} m / s

Answer: 4125

Solution

Net energy absorbed:

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}})

. Substituting values:

E = 6.6 \times 1 0^{- 34} \times 3 \times 1 0^{8} (\frac{1}{500 \times 1 0 ^{- 9}} - \frac{1}{600 \times 1 0 ^{- 9}}) = 6.6 \times 1 0^{- 20}

= \frac{6.6 \times 1 0 ^{- 20}}{1.6 \times 1 0 ^{- 19}}

= 0.4125

= 4125 \times 1 0^{- 4}

eV. Therefore,

n = 4125

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Integer answer

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