Question

De Broglie wavelength ‘$\lambda$ ’ is proportional to

• A. $\frac{1}{\sqrt{E}}$ for photons and $\frac{1}{E}$ for particles
• B. $\frac{1}{E}$ for photons and $\frac{1}{\sqrt{E}}$ for particles
• C. $\frac{1}{E}$ for both photons and particles in motion
• D. $\frac{1}{\sqrt{E}}$ for both photons and particles

Answer 1

The correct option is B $\frac{1}{E}$ for photons and $\frac{1}{\sqrt{E}}$ for particles
For photon
$E=\frac{hc}{\lambda }$

$\lambda =\frac{hc}{E}$

$\therefore \lambda \alpha \frac{1}{E}$

and for partides
$\frac{1}{2}m{v}^2=E$

$mv=\sqrt{2Em}$

$\therefore \lambda =\frac{h}{mv}$

$=\frac{h}{\sqrt{2Em}}$

$\lambda \alpha \frac{1}{\sqrt{E}}$

So, the answer is option (B).