1. Tardigrade
  2. Question
  3. Physics
  4. Photons of energy 7 eV are incident on two metals A and B with work functions 6 eV and 3 eV respectively. The minimum de-Broglie wavelengths of the emitted photoelectrons with maximum energies are λA and λB respectively, where λA / λB is nearly

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Photons of energy $7 \,eV$ are incident on two metals $A$ and $B$ with work functions $6 \,eV$ and $3 \,eV$ 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

KVPYKVPY 2013Dual Nature of Radiation and Matter

Solution:

Given situation is
image
Energy associated with most energised electron is
for metal $A, E_{A} = 7 - 6 = 1\,eV$
and for metal $B, E_{B} = 7 - 3 = 4 \,eV$
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 $