A photon of energy 8,eV is incident on a meta | vedclass

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(C) The work function $W_0$ is given by $W_0 = h
u_0$,where $h = 6.6 	imes 10^{-34}\,Js$ and $
u_0 = 1.6 	imes 10^{15}\,Hz$.$W_0 = (6.6 	imes 10^

Vedclass · Accepted Answer

$1.4$

Vedclass · Answer

(C) The work function $W_0$ is given by $W_0 = h
u_0$,where $h = 6.6 	imes 10^{-34}\,Js$ and $
u_0 = 1.6 	imes 10^{15}\,Hz$.$W_0 = (6.6 	imes 10^{-34}) 	imes (1.6 	imes 10^{15}) = 10.56 	imes 10^{-19}\,J$.To convert this into electron-volts $(eV)$,divide by the charge of an electron $(1.6 	imes 10^{-19}\,C)$:$W_0 = \frac{10.56 	imes 10^{-19}}{1.6 	imes 10^{-19}}\,eV = 6.6\,eV$.According to Einstein's photoelectric equation,$K_{\max} = E - W_0$.Given $E = 8\,eV$ and $W_0 = 6.6\,eV$,we have:$K_{\max} = 8\,eV - 6.6\,eV = 1.4\,eV$.

$A$ photon of energy $8\,eV$ is incident on a metal surface of threshold frequency $1.6 \times 10^{15}\,Hz$. The maximum kinetic energy of the emitted photoelectrons is ...............$eV$ $(h = 6.6 \times 10^{-34}\,Js)$.

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