Question

Photons of wavelength $\lambda$ are incident on the cathode of a photocell. Electrons are emitted from the cathode surface. The de-Broglie wavelength of the emitted electrons (work function is negligible) is
$\left(c = \text{velocity of light},\; h = \text{Planck’s constant},\; m = \text{mass of electron}\right)$

• A. $\sqrt{\dfrac{mc}{2h\lambda}}$
• B. $\sqrt{\dfrac{h\lambda}{2mc}}$
• C. $\sqrt{\dfrac{2h\lambda}{mc}}$
• D. $\sqrt{\dfrac{mh}{\lambda c}}$

Hint:

If work function is negligible, entire photon energy converts into kinetic energy.

Answer 1

The Correct Option is B

Step 1: Energy of incident photon.
\[ E = \frac{hc}{\lambda} \]
Step 2: Kinetic energy of emitted electron.
Since work function is negligible:
\[ K = \frac{hc}{\lambda} \]
Step 3: de-Broglie wavelength of electron.
\[ \lambda_e = \frac{h}{\sqrt{2mK}} \]
Step 4: Substituting $K$.
\[ \lambda_e = \frac{h}{\sqrt{2m\left(\frac{hc}{\lambda}\right)}} = \sqrt{\frac{h\lambda}{2mc}} \]
Step 5: Conclusion.
The de-Broglie wavelength is $\sqrt{\dfrac{h\lambda}{2mc}}$.