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Question

(C) According to Einstein's photoelectric equation: $h
u = K_{max} + \phi_0$, where $K_{max}$ is the maximum kinetic energy and $\phi_0$ is the work

Vedclass · Accepted Answer

Cut-off voltage becomes more than double

Vedclass · Answer

(C) According to Einstein's photoelectric equation: $h
u = K_{max} + \phi_0$, where $K_{max}$ is the maximum kinetic energy and $\phi_0$ is the work function.Initially, $K_1 = h
u - \phi_0$.When the frequency is doubled to $2
u$, the new maximum kinetic energy $K_2$ is:$K_2 = h(2
u) - \phi_0 = 2h
u - \phi_0$.Since $h
u = K_1 + \phi_0$, we substitute this into the equation for $K_2$:$K_2 = 2(K_1 + \phi_0) - \phi_0 = 2K_1 + \phi_0$.Since the stopping potential $V_s$ is related to kinetic energy by $K_{max} = eV_s$, we have $eV_{s2} = 2eV_{s1} + \phi_0$, which implies $V_{s2} = 2V_{s1} + \frac{\phi_0}{e}$.Thus, the stopping potential (cut-off voltage) becomes more than double the initial value.

If the frequency of light falling on a photosensitive material doubles, which of the following is true?

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