The threshold frequency of a metal is 10^{15} | vedclass

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Question

(B) Given threshold frequency, $
u_0 = 1.0 	imes 10^{15} \ s^{-1}$.According to Einstein's photoelectric equation, the maximum kinetic energy $(K.E.

Vedclass · Accepted Answer

$1:2$

Vedclass · Answer

(B) Given threshold frequency, $
u_0 = 1.0 	imes 10^{15} \ s^{-1}$.According to Einstein's photoelectric equation, the maximum kinetic energy $(K.E._{max})$ is given by $K.E._{max} = h(
u - 
u_0)$, where $
u$ is the frequency of incident radiation.For $
u_1 = 1.5 	imes 10^{15} \ s^{-1}$:$K.E._1 = h(1.5 	imes 10^{15} - 1.0 	imes 10^{15}) = h(0.5 	imes 10^{15})$.For $
u_2 = 2.0 	imes 10^{15} \ s^{-1}$:$K.E._2 = h(2.0 	imes 10^{15} - 1.0 	imes 10^{15}) = h(1.0 	imes 10^{15})$.The ratio of maximum kinetic energies is:$\frac{K.E._1}{K.E._2} = \frac{h(0.5 	imes 10^{15})}{h(1.0 	imes 10^{15})} = \frac{0.5}{1.0} = 1:2$.

The threshold frequency of a metal is $10^{15} \ s^{-1}$. The ratio of maximum kinetic energies of the photoelectrons, when the metal is irradiated with radiations of frequencies $1.5 \times 10^{15} \ s^{-1}$ and $2.0 \times 10^{15} \ s^{-1}$ respectively, is

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