In the experiment of the photoelectric effect | vedclass

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Question

(C) According to Einstein's photoelectric equation, the maximum kinetic energy $(K_{max})$ is given by: $K_{max} = hv - \Phi$, where $h$ is Planck's c

Vedclass · Accepted Answer

$v_2 < 2v_1$

Vedclass · Answer

(C) According to Einstein's photoelectric equation, the maximum kinetic energy $(K_{max})$ is given by: $K_{max} = hv - \Phi$, where $h$ is Planck's constant, $v$ is the frequency of incident light, and $\Phi$ is the work function of the metal.For the first case: $K_0 = hv_1 - \Phi$ --- (Equation $1$)For the second case: $2K_0 = hv_2 - \Phi$ --- (Equation $2$)From Equation $1$, we have $\Phi = hv_1 - K_0$. Substituting this into Equation $2$:$2K_0 = hv_2 - (hv_1 - K_0)$$2K_0 = hv_2 - hv_1 + K_0$$K_0 = hv_2 - hv_1$$hv_2 = hv_1 + K_0$Since $K_0 = hv_1 - \Phi$, we substitute $K_0$ back:$hv_2 = hv_1 + (hv_1 - \Phi)$$hv_2 = 2hv_1 - \Phi$$v_2 = 2v_1 - \frac{\Phi}{h}$Since $\Phi > 0$, it follows that $v_2 < 2v_1$.

In the experiment of the photoelectric effect, if the frequency of incident light is $v_1$, the maximum kinetic energy $(K.E.)$ of photoelectrons is $K_0$. If the frequency of light is $v_2$, the maximum $K.E.$ of photoelectrons is $2K_0$. Which of the following relations is correct?

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