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Question

(B) According to Einstein's photoelectric equation: $K_{max} = h
u - \phi_0$,where $\phi_0 = h
u_0$.For frequency $
u_1 = 2
u_0$,the maximum kinet

Vedclass · Accepted Answer

$2\sqrt{2} 	imes 10^6 \ m/s$

Vedclass · Answer

(B) According to Einstein's photoelectric equation: $K_{max} = h
u - \phi_0$,where $\phi_0 = h
u_0$.For frequency $
u_1 = 2
u_0$,the maximum kinetic energy is $K_1 = h(2
u_0) - h
u_0 = h
u_0$.Given $v_1 = 2 	imes 10^6 \ m/s$,so $K_1 = \frac{1}{2}mv_1^2 = h
u_0$.For frequency $
u_2 = 3
u_0$,the maximum kinetic energy is $K_2 = h(3
u_0) - h
u_0 = 2h
u_0$.Since $K_2 = 2K_1$,we have $\frac{1}{2}mv_2^2 = 2(\frac{1}{2}mv_1^2)$.Thus,$v_2^2 = 2v_1^2$,which implies $v_2 = \sqrt{2}v_1$.Substituting $v_1 = 2 	imes 10^6 \ m/s$,we get $v_2 = \sqrt{2} 	imes 2 	imes 10^6 = 2\sqrt{2} 	imes 10^6 \ m/s$.

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