JEE PYQ: Atomic Physics Question 43
Question 43 - 2020 (09 Jan Shift 1)
Radiation, with wavelength 6561 A falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{-4}$ T. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to:
(1) 1.1 eV
(2) 0.8 eV
(3) 1.6 eV
(4) 1.8 eV
Show Answer
Answer: (1)
Solution
Using Einstein’s photoelectric equation, $E = \phi_0 + KE_{max}$. $\phi_0 = KE_{max} - E$. $p = \sqrt{2mKE} \Rightarrow KE = \frac{p^2}{2m}$. $r = \frac{p}{eB} \Rightarrow p = reB$. $KE_{max} = \frac{r^2 e^2 B^2}{2m}$. $\phi_0 = \frac{12420}{6561} - \frac{r^2 e^2 B^2}{2m} = 1.89 - 0.79 \approx 1.1$ eV.
BETA
