Question

Let p and E denote the linear momentum and energy of a photon. If the wavelength is decreased,
(a) both p and E increase
(b) p increases and E decreases
(c) p decreases and E increases
(d) both p and E decrease

Answer 1

(a) both p and E increase

From the de-Broglie relation, wavelength,
$\lambda =\frac{h}{p}...\left(1\right)$
$\Rightarrow$ $p=\frac{h}{\lambda }$
Here, h = Planck's constant
p = momentum of electron
It is clear from the above equation that $p\propto \frac{1}{\lambda }$.
Thus, if the wavelength $\left(\lambda \right)$ is decreased, then momentum $\left(p\right)$ will be increase.
Relation between momentum and energy:
$p=\sqrt{2mE}$
Here, E = energy of electron
m = mass of electron
Substituting the value of p in equation (1), we get:
$$\begin{gathered} \lambda =\frac{h}{\sqrt{2mE}} \\ \Rightarrow \sqrt{E}=\frac{h}{\lambda \sqrt{2m}} \\ \Rightarrow E=\frac{h^2}{2m{\lambda }^2} \end{gathered}$$
Thus, on decreasing $\lambda$, the energy will increase.