Question

A photon of energy $10.2eV$ corresponds to light of wavelength ${\lambda }_0.$ Due to an electron transition from $n=2$ to $n=1$ in a hydrogen atom, the light of wavelength $\lambda$ is emitted. If we take into account the recoil of the atom when the photon is emitted, then

• A. $\lambda <{\lambda }_0$
• B. $\lambda >{\lambda }_0$
• C. $\lambda ={\lambda }_0$
• D. None of these

Answer 1

Answer: (B)

Without recoil of the atom, the entire energy is carried away by the photon.
$\frac{hc}{{\lambda }_0}=10.2eV$
With recoil of the atom, energy is shared between photon and atom. So, the energy of the photon is less, hence its wavelength is more.
$\frac{hc}{\lambda }<\frac{hc}{{\lambda }_0}$
$\therefore {\lambda }_0<\lambda$