When photons of energy 4.25 eV strike the surface of a metal 'A', the ejected photoelectrons have ma — Atomic Structure Chemistry Question
Question
When photons of energy 4.25 eV strike the surface of a metal 'A', the ejected photoelectrons have maximum kinetic energy, $T_A$ (in eV) and de-Broglie wavelength, $\lambda_A$. The maximum kinetic energy of photoelectrons liberated from another metal 'B' by photons of energy 4.20 eV is $T_B$ ($= T_A - 1.50$ eV). If the de-Broglie wave length of these photoelectrons is $\lambda_B$ ($= 2\lambda_A$), then
Answer: A,B,C
💡 Solution & Explanation
$\lambda_B = 2\lambda_A \implies p_B = \frac{p_A}{2} \implies T_B = \frac{T_A}{4}$. Given $T_B = T_A - 1.50 \implies \frac{T_A}{4} = T_A - 1.50 \implies T_A = 2.00$ eV. $T_B = 0.50$ eV. Work function $W_A = 4.25 - 2.00 = 2.25$ eV. Work function $W_B = 4.20 - 0.50 = 3.70$ eV. Therefore, correct answer is A, B, C.
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