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(B) According to Einstein's photoelectric equation, $K_{max} = h
u - \Phi_0 = \frac{hc}{\lambda} - \Phi_0$. As the wavelength $\lambda$ of the incide

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Velocity of emitted photoelectron increases

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(B) According to Einstein's photoelectric equation, $K_{max} = h
u - \Phi_0 = \frac{hc}{\lambda} - \Phi_0$. As the wavelength $\lambda$ of the incident photon decreases, the energy of the incident photon $(E = \frac{hc}{\lambda})$ increases. Since the work function $\Phi_0$ of the metal surface remains constant, the maximum kinetic energy $K_{max}$ of the emitted photoelectrons increases. Because $K_{max} = \frac{1}{2}mv_{max}^2$, an increase in $K_{max}$ leads to an increase in the maximum velocity $v_{max}$ of the emitted photoelectrons.

When the wavelength of an incident photon is decreased, then:

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