Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively, successively illuminate a metallic surface whose work function is 0.5 eV . Ratio of maximum speeds of emitt

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Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively, successively illuminate a metallic surface whose work function is 0.5 eV . Ratio of maximum speeds of emitted electrons will be

TestHub · Accepted Answer

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is $K_{m a x} = h v - ϕ_{0}$ where $h v$ is the energy of incident photon and $ϕ_{0}$ is the work function. But $K_{m a x} = \frac{1}{2} m v_{m a x}^{2}$ $∴ \frac{1}{2} m v_{m a x}^{2} = h v - ϕ_{0}$ As per question,

Dividing eqn. (i) by eqn. (ii), we get $\frac{v _{m a x_{1}}^{2}}{v _{m a x_{2}}^{2}} = \frac{0.5 eV}{2 eV} = \frac{1}{4} \frac{v _{m a x_{1}}}{v _{m a x_{2}}} = \frac{1}{4} = \frac{1}{2}$

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