Question

7. Let p and E denote the linear momentum and energy of a photon. If the wavelength is decreased,
A) both p and E decrease
B) both p and E increase
C) p increases and E decreases
D)p decreases and E increases​

Answer 1

Answer:

hii mate

F = hc/lamda

pc= E

which implies, P=h/lamda

so both of them will be increase with decreasing lamda..

thank you..

Answer 2

Let p and E denote the linear momentum and energy of a photon. If the wavelength is decreased,  which is both p and E increase.

Explanation:

From the relation of de-Broglie, wavelength, $\lambda=\frac{h}{p}$   …1

$\Rightarrow p=\frac{h}{\lambda}$

Here, h = The constant of Planck  

p = electron’s momentum  

So,$p \propto \frac{1}{\lambda}$

Thus, if there is a decrease in wavelength λ, then there will be an increase in the momentum p.

The relation between energy and momentum:

$p=\sqrt{2 m E}$

where, E = electron’s energy

m = electron’s mass

When equation (1) is substituted in the value of p, we obtain:

$\lambda=\frac{h}{\sqrt{2 m E}}$

$\Rightarrow E=\frac{h^{2}}{2 m \lambda^{2}}$

Thus, there will be an increase in the energy when λ decreases.