Question

Radiations of two different frequencies whose photon energies are $3.4eV$ and $8.2eV$ successive illuminate a metal surface whose work function is $1.8eV.$ The ratio of the maximum speeds of the emitted electrons will be:

• A. $1:1$
• B. $1:2$
• C. $1:3$
• D. $1:4$

Answer 1

The correct option is B $1:2$

Energy, $E_1=3.4eV,E_2=8.2eV$

Work function, $W=1.8eV$

From photoelectric law,

$\frac{1}{2}m{v_{max}}^2=E-W\therefore \frac{1}{2}m{v_{1_{max}}}^2=3.4-1.8=1.6eVand\frac{1}{2}m{v_{2_{max}}}^2=8.2-1.8=6.4eV\therefore \frac{\frac{1}{2}m{v_{1_{max}}}^2}{\frac{1}{2}m{v_{2_{max}}}^2}=\frac{1.6}{6.4}\Rightarrow \frac{v_{1_{max}}}{v_{2_{max}}}=\sqrt{\frac{1.6}{6.4}}=\frac{1}{2}$