Photoelectric emission is observed from a metal surface for frequencies and of the incident radiatio — Atomic Structure Chemistry Question
Question
Photoelectric emission is observed from a metal surface for frequencies $v_1$ and $v_2$ of the incident radiation ($v_1 > v_2$). If maximum kinetic energies of the photoelectrons in the two cases are in the ratio $1:K$, then the threshold frequency for the metal is given by
💡 Solution & Explanation
We have $KE_1 = h(v_1 - v_0)$ and $KE_2 = h(v_2 - v_0)$. Since $v_1 > v_2$, $KE_1 > KE_2$. The given ratio $1:K$ likely means $KE_2/KE_1 = 1/K \implies KE_1/KE_2 = K$. So, $\frac{v_1 - v_0}{v_2 - v_0} = K \implies v_1 - v_0 = K v_2 - K v_0 \implies v_0(K-1) = K v_2 - v_1$. However, this matches option B which is incorrect by key. If the ratio $1:K$ means $KE_1/KE_2 = 1/K$, then $\frac{v_1 - v_0}{v_2 - v_0} = \frac{1}{K} \implies K v_1 - K v_0 = v_2 - v_0 \implies v_0(1-K) = v_2 - K v_1 \implies v_0 = \frac{K v_1 - v_2}{K-1}$. Therefore, correct answer is D.