(A) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi$,where $K_{max} = \frac{1}{2}mv^2$.For wavelength $\lambda$:

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(A) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi$,where $K_{max} = \frac{1}{2}mv^2$.For wavelength $\lambda$: $\frac{1}{2}mV^2 = \frac{hc}{\lambda} - \phi$ --- $(1)$For wavelength $\lambda' = \frac{2\lambda}{3}$: $\frac{1}{2}mv'^2 = \frac{hc}{2\lambda/3} - \phi = \frac{3hc}{2\lambda} - \phi$ --- $(2)$From $(1)$,$\frac{hc}{\lambda} = \frac{1}{2}mV^2 + \phi$.Substitute this into $(2)$: $\frac{1}{2}mv'^2 = \frac{3}{2}(\frac{1}{2}mV^2 + \phi) - \phi = \frac{3}{4}mV^2 + \frac{1}{2}\phi$.Since $\phi > 0$,$\frac{1}{2}mv'^2 > \frac{3}{4}mV^2$.$v'^2 > \frac{3}{2}V^2$,which implies $v' > \sqrt{\frac{3}{2}}V$.

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

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When a light of wavelength $\lambda$ falls on the emitter of a photocell,the maximum speed of emitted photoelectrons is $V$. If the incident wavelength is changed to $2\lambda / 3$,the maximum speed of the emitted photoelectrons will be:

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A
Greater than $\sqrt{3/2} \,V$
B
Less than $\sqrt{3/2} \,V$
C
Equal to $\sqrt{3/2} \,V$
D
Equal to $V$

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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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When a light of wavelength $\lambda$ falls on the emitter of a photocell,the maximum speed of emitted photoelectrons is $V$. If the incident wavelength is changed to $2\lambda / 3$,the maximum speed of the emitted photoelectrons will be:

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