In a photoelectric effect experiment,the kine | vedclass

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Question

(A) Given,kinetic energy of an emitted electron $(KE) = 1.986 	imes 10^{-19} \ J$. Frequency of radiation $(
u) = 1.0 	imes 10^{15} \ s^{-1}$. Plan

Vedclass · Accepted Answer

$7.0 	imes 10^{14}$

Vedclass · Answer

(A) Given,kinetic energy of an emitted electron $(KE) = 1.986 	imes 10^{-19} \ J$. Frequency of radiation $(
u) = 1.0 	imes 10^{15} \ s^{-1}$. Planck's constant $(h) = 6.62 	imes 10^{-34} \ J \ s$. According to the photoelectric effect equation: $h
u = h
u_0 + KE$,where $
u_0$ is the threshold frequency. Rearranging for $
u_0$: $h
u_0 = h
u - KE$. $
u_0 = 
u - \frac{KE}{h}$. Substituting the values: $
u_0 = 1.0 	imes 10^{15} \ s^{-1} - \frac{1.986 	imes 10^{-19} \ J}{6.62 	imes 10^{-34} \ J \ s}$. $
u_0 = 1.0 	imes 10^{15} \ s^{-1} - 0.3 	imes 10^{15} \ s^{-1}$. $
u_0 = 0.7 	imes 10^{15} \ s^{-1} = 7.0 	imes 10^{14} \ s^{-1}$.

In a photoelectric effect experiment,the kinetic energy of an emitted electron is $1.986 \times 10^{-19} \ J$,when a radiation of frequency $1.0 \times 10^{15} \ s^{-1}$ hits the metal. What is the threshold frequency of the metal (in $s^{-1}$)? (Planck's constant $= 6.62 \times 10^{-34} \ J \ s$)

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