Q 12-12-067JEE MainJEE Main 2023 (13 Apr, Shift 2)Medium
An atom absorbs a photon of wavelength $500\ \text{nm}$ and emits another photon of wavelength $600\ \text{nm}$. The net energy absorbed by the atom in this process is $n\times10^{-4}\ \text{eV}$. The value of $n$ is ______. [Assume the atom to be stationary during the absorption and emission process] (Take $h=6.6\times10^{-34}\ \text{J s}$ and $c=3\times10^8\ \text{m s}^{-1}$)
Numerical value type. Enter your answer.
Answer: 4125
$hc=1.98\times10^{-25}\ \text{J m}=1237.5\ \text{eV nm}$.
$$\Delta E=1237.5\left(\frac1{500}-\frac1{600}\right)=\frac{1237.5}{3000}=0.4125\ \text{eV}=4125\times10^{-4}\ \text{eV}$$
Solution by Sreeraj P, M.Sc Physics