In the experiment of P.E.E.,the KE_{max} of a | vedclass

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(A) According to Einstein's photoelectric equation,$hf = \phi_0 + KE_{max}$.For the $1^{	ext{st}}$ case: $hf = \phi_0 + K_0$  --- $(1)$For the $2^{

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$\left( \frac{n_2 - n_1}{n_1 - 1} \right) K_0$

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(A) According to Einstein's photoelectric equation,$hf = \phi_0 + KE_{max}$.For the $1^{	ext{st}}$ case: $hf = \phi_0 + K_0$  --- $(1)$For the $2^{	ext{nd}}$ case,the frequency is increased by a factor of $n_1$,so the new frequency is $n_1f$. The new $KE_{max}$ is $n_2K_0$:$n_1hf = \phi_0 + n_2K_0$  --- $(2)$Substitute $hf$ from equation $(1)$ into equation $(2)$:$n_1(\phi_0 + K_0) = \phi_0 + n_2K_0$Expand the equation:$n_1\phi_0 + n_1K_0 = \phi_0 + n_2K_0$Rearrange to solve for $\phi_0$:$n_1\phi_0 - \phi_0 = n_2K_0 - n_1K_0$$\phi_0(n_1 - 1) = K_0(n_2 - n_1)$Therefore,the work function is:$\phi_0 = \left( \frac{n_2 - n_1}{n_1 - 1} ight) K_0$

In the experiment of $P.E.E.$,the $KE_{max}$ of an electron is $K_0$. If the frequency is increased by a factor of $n_1$,then the $KE_{max}$ becomes $n_2K_0$. Find the work function.

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