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

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(A) Given:Energy of incident photon,$E = 8\,eV$.Threshold frequency,$
u_0 = 1.6 	imes 10^{15}\,Hz$.Planck's constant,$h = 6.6 	imes 10^{-34}\,J\cdo

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(A) Given:Energy of incident photon,$E = 8\,eV$.Threshold frequency,$
u_0 = 1.6 	imes 10^{15}\,Hz$.Planck's constant,$h = 6.6 	imes 10^{-34}\,J\cdot s$.Conversion factor,$1\,eV = 1.6 	imes 10^{-19}\,J$.The work function $\Phi$ is given by $\Phi = h
u_0$.$\Phi = (6.6 	imes 10^{-34}\,J\cdot s) 	imes (1.6 	imes 10^{15}\,Hz) = 10.56 	imes 10^{-19}\,J$.Converting the work function into $eV$:$\Phi = \frac{10.56 	imes 10^{-19}\,J}{1.6 	imes 10^{-19}\,J/eV} = 6.6\,eV$.According to Einstein's photoelectric equation,the maximum kinetic energy $KE_{\max}$ is:$KE_{\max} = E - \Phi$.$KE_{\max} = 8\,eV - 6.6\,eV = 1.4\,eV$.

$A$ 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 photoelectrons emitted is .......... $eV$.
(Take $h = 6.6 \times 10^{-34}\,J\cdot s$; $1\,eV = 1.6 \times 10^{-19}\,J$)

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$A$ 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 photoelectrons emitted is .......... $eV$.(Take $h = 6.6 \times 10^{-34}\,J\cdot s$; $1\,eV = 1.6 \times 10^{-19}\,J$)