Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The electric field at the point associated with a light wave is given by $E=200\left[\sin \left(6 \times 10^{15}\right) t +\sin \left(9 \times 10^{15}\right) t \right] Vm ^{-1}$ Given : $h =4.14 \times 10^{-15} eVs$ If this light falls on a metal surface having a work function of $2.50\, eV$, the maximum kinetic energy of the photoelectrons will be :

JEE MainJEE Main 2022Dual Nature of Radiation and Matter

Solution:

For maximum KE we will take higher frequency
$\left( f =\frac{9 \times 10^{15}}{2 \pi} Hz \right)$
$K _{\max }= hf -\phi$
$=\frac{9 \times 10^{15} \times 4.14 \times 10^{-15}}{2 \pi}-2.50$
$3.43\, eV \text { nearest is } 3.42\, eV$