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

Answer: (D)

Given situation is

Energy associated with most energised electron is
for metal $A,E_A=7-6=1eV$
and for metal $B,E_B=7-3=4eV$
Now, de-Broglie wavelength associated with these electrons is
$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mE}}$
$\therefore \frac{{\lambda }_A}{{\lambda }_B}=\frac{\sqrt{E_B}}{\sqrt{E_A}}=\sqrt{\frac{4}{1}}$
$\Rightarrow \frac{{\lambda }_A}{{\lambda }_B}=2:1=2.0$