JEE PYQ: Atomic Physics Question 33
Question 33 - 2020 (04 Sep Shift 2)
The surface of a metal is illuminated alternately with photons of energies $E_1 = 4$ eV and $E_2 = 2.5$ eV respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is 2. The work function of the metal in (eV) is ____.
Show Answer
Answer: 2
Solution
Energy of photon = Kinetic energy of photoelectrons + Work function. $\frac{1}{2}mv_1^2 = 4 - \phi_0$ …(i). $\frac{1}{2}mv_2^2 = 2.5 - \phi_0$ …(ii). $(2)^2 = \frac{4 - \phi_0}{2.5 - \phi_0}$. $10 - 4\phi_0 = 4 - \phi_0$. $\phi_0 = 2$ eV.
BETA
