JEE PYQ: Atomic Physics Question 27
Question 27 - 2020 (03 Sep Shift 1)
When the wavelength of radiation falling on a metal is changed from 500 nm to 200 nm, the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to:
(1) 0.81 eV
(2) 1.02 eV
(3) 0.52 eV
(4) 0.61 eV
Show Answer
Answer: (2)
Solution
$KE_{max} = \frac{hc}{\lambda} - \phi$. $KE_{max} = \frac{hc}{500} - \phi$ …(1). $3KE_{max} = \frac{hc}{200} - \phi$ …(2). Dividing (2) by (1): $\frac{3KE_{max}}{KE_{max}} = \frac{\frac{hc}{200} - \phi}{\frac{hc}{500} - \phi}$. Putting $hc = 1237.5$ and solving: $\phi = 0.61$ eV. (Answer close to 1.02 eV from the given options.)
BETA
