(B) According to Einstein's photoelectric equation: $K_{max} = h\nu - \phi_0$,where $\phi_0 = h\nu_0$.For frequency $\nu_1 = 2\nu_0$,the maximum kinet

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(B) According to Einstein's photoelectric equation: $K_{max} = h\nu - \phi_0$,where $\phi_0 = h\nu_0$.For frequency $\nu_1 = 2\nu_0$,the maximum kinetic energy is $K_1 = h(2\nu_0) - h\nu_0 = h\nu_0$.Given $v_1 = 2 \times 10^6 \ m/s$,so $K_1 = \frac{1}{2}mv_1^2 = h\nu_0$.For frequency $\nu_2 = 3\nu_0$,the maximum kinetic energy is $K_2 = h(3\nu_0) - h\nu_0 = 2h\nu_0$.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 $v_1 = 2 \times 10^6 \ m/s$,we get $v_2 = \sqrt{2} \times 2 \times 10^6 = 2\sqrt{2} \times 10^6 \ m/s$.

Class 12 Physics Dual Nature of Radiation and matter Einstein's Photoelectric Equation and Energy Quantum of Radiation

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The threshold frequency for a certain metal is $v_0$. When radiation of frequency $2v_0$ is incident on this metal surface,the maximum velocity of the emitted photoelectrons is $2 \times 10^6 \ m/s$. If radiation of frequency $3v_0$ is incident on the same metal surface,the maximum velocity of the emitted photoelectrons is . . . . . .

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A
$\sqrt{2} \times 10^6 \ m/s$
B
$2\sqrt{2} \times 10^6 \ m/s$
C
$4 \times 10^6 \ m/s$
D
$5 \times 10^6 \ m/s$

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Dual Nature of Radiation and matter

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Einstein's Photoelectric Equation and Energy Quantum of Radiation

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