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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the emitted photoelectrons is given by: $K_{max} = E - \Phi

Vedclass · Accepted Answer

$4.31 	imes 10^{-25} kgms^{-1}$

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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the emitted photoelectrons is given by: $K_{max} = E - \Phi$,where $E = 5 eV$ is the energy of the incident photon and $\Phi = 4.36 eV$ is the work function of the metal. $K_{max} = 5 eV - 4.36 eV = 0.64 eV$. Converting this energy into Joules: $K_{max} = 0.64 	imes 1.6 	imes 10^{-19} J = 1.024 	imes 10^{-19} J$. The relationship between maximum kinetic energy and maximum momentum $p$ is $K_{max} = \frac{p^2}{2m}$,where $m$ is the mass of an electron $(9.1 	imes 10^{-31} kg)$. $p = \sqrt{2m K_{max}} = \sqrt{2 	imes 9.1 	imes 10^{-31} 	imes 1.024 	imes 10^{-19}}$. $p = \sqrt{18.6368 	imes 10^{-50}} \approx 4.31 	imes 10^{-25} kgms^{-1}$.

When photons of energy $5 eV$ are incident on a metal surface with a work function of $4.36 eV$,photoelectrons are emitted. What is the maximum momentum of the emitted photoelectrons?

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