Light of two different frequencies whose photons have energies $1.3 \mathrm{eV}$ and $2.8 \mathrm{eV}$…
Light of two different frequencies whose photons have energies $1.3 \mathrm{eV}$ and $2.8 \mathrm{eV}$ respectively, successfully illuminate a metallic surface whose work function is $0.8 \mathrm{eV}$. The ratio of maximum speeds of emitted electrons will be
$1: 4$
$1: 2$
$1: 3$
1:5
Solution
For maximum speed of the photo electrons, $\frac{1}{2} m v^2=E_{\mathrm{P}}-\phi$ where $\phi=0.8 \mathrm{eV}$ is the work function of the metal. Energy of photon $E_1=1.3 \mathrm{eV}$
$\therefore \frac{1}{2} m v_1^2=1.3-0.8 \mathrm{eV}$
Ratio of maximum speeds of the electrons can obtained by taking ratio of equation (1) and (2),
$\therefore \frac{v_1^2}{v_2^2}=\frac{0.5}{2}=\frac{1}{4}$
Or, $\frac{v_1}{v_2}=\frac{1}{2}$
Ratio of maximum speeds of the electrons can obtained by taking ratio of equation (1) and (2),
$\therefore \frac{v_1^2}{v_2^2}=\frac{0.5}{2}=\frac{1}{4}$
Or, $\frac{v_1}{v_2}=\frac{1}{2}$