If the work function for a certain metal is 3 | vedclass

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(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by:$K_{\max} = h
u - \phi_0$Where $h$ is Planck's c

Vedclass · Accepted Answer

$2.1 	imes 10^{-19}$

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(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by:$K_{\max} = h
u - \phi_0$Where $h$ is Planck's constant $(6.63 	imes 10^{-34} \ J \cdot s)$,$
u$ is the frequency of incident light,and $\phi_0$ is the work function.Given: $\phi_0 = 3.2 	imes 10^{-19} \ J$ and $
u = 8 	imes 10^{14} \ Hz$.First,calculate the energy of the incident photon $(E = h
u)$:$E = (6.63 	imes 10^{-34} \ J \cdot s) 	imes (8 	imes 10^{14} \ Hz) = 53.04 	imes 10^{-20} \ J = 5.304 	imes 10^{-19} \ J$.Now,substitute the values into the equation:$K_{\max} = 5.304 	imes 10^{-19} \ J - 3.2 	imes 10^{-19} \ J$$K_{\max} = 2.104 	imes 10^{-19} \ J \approx 2.1 	imes 10^{-19} \ J$.

If the work function for a certain metal is $3.2 \times 10^{-19} \ J$ and it is illuminated with light of frequency $8 \times 10^{14} \ Hz$,the maximum kinetic energy of the photoelectrons would be in joule:

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