Question

The threshold frequency of a metal is 1 x 10^15 Hz. The ratio of maximum kinetic energies of the photoelectrons when the metal is irradiated with radiations of frequencies 1.5x 10¹5 and 2x10^15 Hz respectively would be.​

Answer 1

Answer:

K.E.(max) = h ( ν - ν₀)

Ratio of kinetic energies = $\frac{K.E.(1)}{K.E.(2)} = \frac{h(v1-v0)}{h(v2-v0)}\\=\frac{v1-v0}{v2-v0}\\=\frac{1.5*10^{15}-1 *10^{15}}{2*10^{15}-1*10^{15}} \\=\frac{0.5}{1} =\frac{1}{2}$