Question

A metal surface is irradiated by radiations whose wavelength is $\lambda$. If the work function of the metal surface is negligibly small, then the de-Broglie wavelength of emitted electrons will be

• A. $\left[\dfrac{h\lambda}{2mc}\right]^{1/2}$
• B. $\left[\dfrac{h\lambda}{2mc}\right]^{1/3}$
• C. $\left[\dfrac{2hc}{\lambda}\right]^{1/2}$
• D. $\left[\dfrac{hc}{2m\lambda}\right]^{1/3}$

Hint:

When work function is negligible, photon energy directly converts into kinetic energy of emitted electrons.

Answer 1

The Correct Option is A

Step 1: Write expression for photon energy.
The energy of incident radiation is given by: \[ E = \dfrac{hc}{\lambda} \] Step 2: Use photoelectric effect condition.
Since the work function is negligibly small, the entire photon energy is converted into kinetic energy of the electron: \[ \dfrac{1}{2}mv^2 = \dfrac{hc}{\lambda} \] Step 3: Express momentum of emitted electron.
\[ p = \sqrt{2m \times \dfrac{hc}{\lambda}} \] Step 4: Use de-Broglie wavelength relation.
\[ \lambda_d = \dfrac{h}{p} \] Step 5: Substitute momentum and simplify.
\[ \lambda_d = \left[\dfrac{h\lambda}{2mc}\right]^{1/2} \]