JEE PYQ: Atomic Physics Question 39
Question 39 - 2020 (08 Jan Shift 1)
When photon of energy 4.0 eV strikes the surface of a metal $A$, the ejected photoelectrons have maximum kinetic energy $T_A$ eV and de-Broglie wavelength $\lambda_A$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by photon of energy 4.50 eV is $T_B = (T_A - 1.5)$ eV. If the de-Broglie wavelength of these photoelectrons $\lambda_B = 2\lambda_A$, then the work function of metal B is:
(1) 4 eV
(2) 2 eV
(3) 1.5 eV
(4) 3 eV
Show Answer
Answer: (1)
Solution
$\lambda = \frac{h}{\sqrt{2mKE}}$, $\lambda \propto \frac{1}{\sqrt{KE}}$. $\frac{\lambda_A}{\lambda_B} = \sqrt{\frac{KE_B}{KE_A}} = \sqrt{\frac{T_A - 1.5}{T_A}}$ (as given). Also $\frac{\lambda_A}{\lambda_B} = \frac{1}{2}$. So $KE_B = T_A - 1.5 = 2 - 1.5 = 0.5$ eV, $T_A = 2$ eV. Work function of metal B: $\phi_B = E_B - KE_B = 4.5 - 0.5 = 4$ eV.
BETA
