Question

Photons of energy $7eV$ are incident on two metals A and B with work functions $6eV$ and $3eV$ respectively. The minimum de Broglie wavelengths of the emitted photoelectrons with maximum energies are ${\lambda }_A$ and ${\lambda }_B$ , respectively where ${\lambda }_A/{\lambda }_B$ is nearly :

• A. $0.5$
• B. $1.4$
• C. $4.0$
• D. $2.0$

Answer 1

The correct option is D $2.0$

Kinetic energy of photo electrons $K=E_{photons}-\varphi$

where $\varphi$ is the work function of the metal and $E_{photon}$ is the energy of the photon.

For metal A : $K_A=7-6=1$ eV

For metal B : $K_B=7-3=4$ eV

de Broglie wavelength $\lambda =\frac{h}{\sqrt{2mK}}$

$⟹$ $\frac{{\lambda }_A}{{\lambda }_B}=\sqrt{\frac{K_B}{K_A}}$

Or $\frac{{\lambda }_A}{{\lambda }_B}=\sqrt{\frac{4}{1}}=2.0$