Discussion Forum

When the momentum of a photon is changed by an amount p' then the corresponding change in the de-Broglie wavelength is found to be 0.20%. Then the original moment of the photon was

Answer:

Explanation:

As, we know de-Broglic  wavelength

 $\lambda=\frac{h}{p}$

$\therefore$     $\lambda\propto\frac{1}{p}$

$\Rightarrow \frac{\triangle p}{p}=-\frac{\triangle \lambda}{\lambda}\therefore |\frac{\triangle p}{p}|=|\frac{\triangle \lambda}{\lambda}|$

$\Rightarrow \frac{p'}{p}=\frac{0.20}{100}=\frac{1}{500}$

 or, p=500 p'