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

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$v_1^2 - v_2^2 = \frac{2h}{m}(f_1 - f_2)$

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(A) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photo-electrons is given by $K_{max} = hf - \phi$,where $\phi = hf_0$ is the work function.For the first photo-cathode: $\frac{1}{2}mv_1^2 = hf_1 - hf_0$ (Equation $1$)For the second photo-cathode: $\frac{1}{2}mv_2^2 = hf_2 - hf_0$ (Equation $2$)Subtracting Equation $2$ from Equation $1$:$\frac{1}{2}mv_1^2 - \frac{1}{2}mv_2^2 = (hf_1 - hf_0) - (hf_2 - hf_0)$$\frac{1}{2}m(v_1^2 - v_2^2) = h(f_1 - f_2)$$v_1^2 - v_2^2 = \frac{2h}{m}(f_1 - f_2)$

Two identical photo-cathodes receive light of frequencies $f_1$ and $f_2$. If the velocities of the photo-electrons (of mass $m$) coming out are $v_1$ and $v_2$ respectively,then:

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