If the maximum kinetic energy of photoelectro | vedclass

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Question

(C) According to the photoelectric effect equation: $h
u = KE_{max} + h
u_0$ Where $h$ is Planck's constant $(6.63 	imes 10^{-34} \ J \cdot s)$,$

Vedclass · Accepted Answer

$3 	imes 10^{14} \ s^{-1}$

Vedclass · Answer

(C) According to the photoelectric effect equation: $h
u = KE_{max} + h
u_0$ Where $h$ is Planck's constant $(6.63 	imes 10^{-34} \ J \cdot s)$,$
u$ is the incident frequency,$KE_{max}$ is the maximum kinetic energy,and $
u_0$ is the threshold frequency. Rearranging for $
u_0$: $h
u_0 = h
u - KE_{max}$ $
u_0 = 
u - \frac{KE_{max}}{h}$ Substituting the given values: $
u_0 = (4 	imes 10^{14} \ s^{-1}) - \frac{6.63 	imes 10^{-20} \ J}{6.63 	imes 10^{-34} \ J \cdot s}$ $
u_0 = 4 	imes 10^{14} \ s^{-1} - 1 	imes 10^{14} \ s^{-1}$ $
u_0 = 3 	imes 10^{14} \ s^{-1}$

If the maximum kinetic energy of photoelectrons ejected from a metal surface when it is irradiated with a radiation of frequency $4 \times 10^{14} \ s^{-1}$ is $6.63 \times 10^{-20} \ J$,then the threshold frequency of the metal is

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