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

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$4 \sqrt{3} 	imes 10^6 \,m/s$

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(C) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ is given by $K_{max} = E - \phi$, where $E$ is the incident photon energy and $\phi$ is the work function.For the first case: $E_1 = 3\phi$. Thus, $K_{max1} = 3\phi - \phi = 2\phi$.We know $K_{max} = \frac{1}{2}mv^2$, so $\frac{1}{2}mv_1^2 = 2\phi$ --- $(1)$.For the second case: $E_2 = 7\phi$. Thus, $K_{max2} = 7\phi - \phi = 6\phi$.So, $\frac{1}{2}mv_2^2 = 6\phi$ --- $(2)$.Dividing equation $(2)$ by equation $(1)$: $\frac{v_2^2}{v_1^2} = \frac{6\phi}{2\phi} = 3$.Therefore, $v_2 = v_1 \sqrt{3}$.Given $v_1 = 4 	imes 10^6 \,m/s$, we get $v_2 = 4 \sqrt{3} 	imes 10^6 \,m/s$.

$A$ photosensitive surface has work function $\phi$. If a photon of energy $3 \phi$ falls on this surface, the electron comes out with a maximum velocity of $4 \times 10^6 \,m/s$. When the photon energy is increased to $7 \phi$, the maximum velocity of the photoelectron will be:

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