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Question

(D) According to Einstein's photoelectric equation, $K_{max} = h
u - \phi$, where $K_{max}$ is the maximum kinetic energy of emitted photoelectrons,

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$D$

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(D) According to Einstein's photoelectric equation, $K_{max} = h
u - \phi$, where $K_{max}$ is the maximum kinetic energy of emitted photoelectrons, $h$ is Planck's constant, $
u$ is the frequency of incident radiation, and $\phi$ is the work function of the metal.Increasing the frequency $(
u)$ of the incident radiation increases the maximum kinetic energy $(K_{max})$ of the emitted photoelectrons. Consequently, the stopping potential $(V_s)$, which is defined as $V_s = K_{max}/e$, becomes more negative (shifts to the left on the potential axis).Intensity of incident radiation is directly proportional to the number of photons striking the surface per unit time, which in turn determines the saturation photocurrent. Reducing the intensity decreases the number of photoelectrons emitted per unit time, thereby reducing the saturation photocurrent.Comparing the given curves with the broken line curve: Curve $D$ shows a more negative stopping potential (due to increased frequency) and a lower saturation current (due to decreased intensity). Therefore, curve $D$ represents the new situation.

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