Question

A photon of energy $10.2 \, eV$ 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.
$\frac{h c}{\lambda _{0}}=10.2eV$
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.
$\frac{h c}{\lambda } < \frac{h c}{\lambda _{0}}$
$\therefore \lambda _{0} < \lambda $