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Question

(D) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K_{\max} = E - W_0$,where $E$ is the energy of the incident

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$8 	imes 10^6 \, m/s$

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K_{\max} = E - W_0$,where $E$ is the energy of the incident photon and $W_0$ is the work function.For the first case: $E_1 = 2h
u_0$ and $W_0 = h
u_0$.$\frac{1}{2}mv_1^2 = 2h
u_0 - h
u_0 = h
u_0$ ... $(i)$For the second case: $E_2 = 5h
u_0$ and $W_0 = h
u_0$.$\frac{1}{2}mv_2^2 = 5h
u_0 - h
u_0 = 4h
u_0$ ... $(ii)$Dividing equation $(ii)$ by $(i)$:$\frac{\frac{1}{2}mv_2^2}{\frac{1}{2}mv_1^2} = \frac{4h
u_0}{h
u_0}$$\left(\frac{v_2}{v_1}ight)^2 = 4$$\frac{v_2}{v_1} = 2$Given $v_1 = 4 	imes 10^6 \, m/s$,we find:$v_2 = 2 	imes v_1 = 2 	imes (4 	imes 10^6 \, m/s) = 8 	imes 10^6 \, m/s$.

$A$ photosensitive metallic surface has a work function $h\nu_0$. If photons of energy $2h\nu_0$ fall on this surface,the electrons are emitted with a maximum velocity of $4 \times 10^6 \, m/s$. When the photon energy is increased to $5h\nu_0$,the maximum velocity of the photoelectrons will be:

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