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

Vedclass · Accepted Answer

$1:2$

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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of emitted electrons is given by $K_{max} = E - \Phi$,where $E$ is the energy of the incident photon and $\Phi$ is the work function of the metal.For the first photon with energy $E_1 = 1 	ext{ eV}$:$K_1 = 1 	ext{ eV} - 0.5 	ext{ eV} = 0.5 	ext{ eV}$.For the second photon with energy $E_2 = 2.5 	ext{ eV}$:$K_2 = 2.5 	ext{ eV} - 0.5 	ext{ eV} = 2.0 	ext{ eV}$.Since $K_{max} = \frac{1}{2}mv^2$,the ratio of the kinetic energies is:$\frac{K_1}{K_2} = \frac{0.5}{2.0} = \frac{1}{4}$.Substituting the expression for kinetic energy:$\frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2} = \frac{1}{4} \Rightarrow \frac{v_1^2}{v_2^2} = \frac{1}{4}$.Taking the square root on both sides:$\frac{v_1}{v_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Thus,the ratio of the maximum speeds is $1:2$.

Light of two different frequencies whose photons have energies $1 \text{ eV}$ and $2.5 \text{ eV}$ respectively,successively illuminate a metallic surface whose work function is $0.5 \text{ eV}$. The ratio of the maximum speeds of the emitted electrons will be:

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