Question

Photon A has twice the energy of photon B. Then which option is correct?

• A. Momentum of A is less than momentum of B.
• B. Wavelength of A is less than wavelength of B.
• C. Wavelength of A is equal to wavelength of B.
• D. Wavelength of A is greater than wavelength of B.

Answer 1

The correct option is B Wavelength of A is less than wavelength of B.

Energy of a photon $E=h\nu$
and momentum of photon $p=\frac{h}{\lambda }$
$\Rightarrow E=p\lambda \nu =pc$

Given, $E_A=2E_B$ and $c$ is a constant.
$\Rightarrow p_A=2p_B$
Now, $\lambda \propto \frac{1}{p}$
$\Rightarrow \frac{1}{{\lambda }_A}=\frac{2}{{\lambda }_B}$
$\Rightarrow {\lambda }_A=\frac{{\lambda }_B}{2}$

So option B is correct.