Two photons of energy 2.5 eV and 3.5 eV fall | vedclass

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of an emitted photoelectron is given by $K_{max} = E - \phi_0$

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$1 : \sqrt{2}$

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of an emitted photoelectron is given by $K_{max} = E - \phi_0$,where $E$ is the energy of the incident photon and $\phi_0$ is the work function of the metal.For the first photon with energy $E_1 = 2.5 eV$:$\frac{1}{2} m v_1^2 = E_1 - \phi_0 = 2.5 eV - 1.5 eV = 1.0 eV$ $(i)$For the second photon with energy $E_2 = 3.5 eV$:$\frac{1}{2} m v_2^2 = E_2 - \phi_0 = 3.5 eV - 1.5 eV = 2.0 eV$ (ii)Dividing equation $(i)$ by equation (ii),we get:$\frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2} = \frac{1.0 eV}{2.0 eV}$$\frac{v_1^2}{v_2^2} = \frac{1}{2}$Taking the square root on both sides:$\frac{v_1}{v_2} = \frac{1}{\sqrt{2}}$Thus,the ratio of the maximum velocities is $1 : \sqrt{2}$.

Two photons of energy $2.5 eV$ and $3.5 eV$ fall on a metal surface of work function $1.5 eV$. The ratio of the maximum velocities of the photoelectrons emitted from the metal surface is

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