A hydrogen atom changes its state from $n = 3$ to $n = 2$. Due to recoil, the percentage change in the wavelength of the emitted light is approximately $1\times10^{-n}$. The value of $n$ is ______. [Given $Rhc = 13.6\ \text{eV}$, $hc = 1242\ \text{eV nm}$, $h = 6.6\times10^{-34}\ \text{J s}$, mass of the hydrogen atom $= 1.6\times10^{-27}\ \text{kg}$]
Numerical value type. Enter your answer.
Answer: 7
Energy released: $E = 13.6\left(\dfrac14 - \dfrac19\right) = 1.89\ \text{eV}$.
The photon momentum $p = E/c$ is given to the atom, whose recoil energy is $\dfrac{p^2}{2M} = \dfrac{E^2}{2Mc^2}$. The photon energy is smaller by this amount, so
$$\frac{\Delta\lambda}{\lambda} \approx \frac{E}{2Mc^2}$$
$Mc^2 = 1.6\times10^{-27}\times9\times10^{16} = 1.44\times10^{-10}\ \text{J} = 9\times10^8\ \text{eV}$.
$$\frac{\Delta\lambda}{\lambda}\times100 = \frac{1.89}{1.8\times10^9}\times100 \approx 1\times10^{-7}\ \%$$
So $n = 7$.
Solution by Sreeraj P, M.Sc Physics