Explanation

(c) Energy of photon,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = hv = \frac{hc}{\lambda}\ \ \ $…(i)
Where $h$ is Planck’s constant and $c$ the speed of light.
Multiplying and dividing the RHS of above expression by $m$ where $m$ is mass,we have
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = \frac{hmc}{m\lambda}$
Now, $mc$=p= momentum
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = \frac{hp}{m\lambda}$
Given, $E_{1} = E,\ \ \ \ \ \ \ \ \ \ \ E_{2} = 4E,\ \ \ \ \ \ \ \ \ \ \ \ \ \ p_{1} = p$
$\frac{E_{1}}{E_{2}} = \frac{p_{1}}{p_{2}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ p_{2} = p_{1}\frac{E_{2}}{E_{1}} = p\frac{4E}{E} = 4p$
Hence, momentum increases by a factor of 4.