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(B) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K_{max} = \frac{1}{2} m v^2 = E_p

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$1: 2$

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K_{max} = \frac{1}{2} m v^2 = E_p - \phi$,where $E_p$ is the energy of the incident photon and $\phi$ is the work function of the metal.Given $\phi = 0.8 eV$.For the first photon with energy $E_1 = 1.3 eV$:$\frac{1}{2} m v_1^2 = 1.3 eV - 0.8 eV = 0.5 eV$  --- $(1)$For the second photon with energy $E_2 = 2.8 eV$:$\frac{1}{2} m v_2^2 = 2.8 eV - 0.8 eV = 2.0 eV$  --- $(2)$Dividing equation $(1)$ by equation $(2)$:$\frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2} = \frac{0.5 eV}{2.0 eV}$$\frac{v_1^2}{v_2^2} = \frac{0.5}{2.0} = \frac{1}{4}$Taking the square root on both sides:$\frac{v_1}{v_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}$Therefore,the ratio of maximum speeds is $1: 2$.

Light of two different frequencies whose photons have energies $1.3 eV$ and $2.8 eV$ respectively,successfully illuminate a metallic surface whose work function is $0.8 eV$. The ratio of maximum speeds of emitted electrons will be

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