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Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum speeds of emitted electrons will be
A
1 : 5
B
1 : 4
C
1 : 2
D
1 : 1
Detailed Solution
Einstein's equation: $\frac{1}{2}mv_{max}^2 = h\nu - \phi_0$
First light: $KE_1 = 1 - 0.5 = 0.5$ eV
Second light: $KE_2 = 2.5 - 0.5 = 2.0$ eV
Since $KE \propto v^2$: $\frac{v_1}{v_2} = \sqrt{\frac{KE_1}{KE_2}} = \sqrt{\frac{0.5}{2.0}} = \sqrt{\frac{1}{4}}$
$\frac{v_1}{v_2} = \frac{1}{2}$, i.e., 1 : 2
First light: $KE_1 = 1 - 0.5 = 0.5$ eV
Second light: $KE_2 = 2.5 - 0.5 = 2.0$ eV
Since $KE \propto v^2$: $\frac{v_1}{v_2} = \sqrt{\frac{KE_1}{KE_2}} = \sqrt{\frac{0.5}{2.0}} = \sqrt{\frac{1}{4}}$
$\frac{v_1}{v_2} = \frac{1}{2}$, i.e., 1 : 2