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JEE MAIN 2019
09-01-19
Question
The magnetic field associated with a light wave is given, at the origin, by $B=B_0\left[\sin \left(3.14 \times 10^7\right) c t+\sin (6.28 \times\right. \left.10^7\right) \mathrm{ct}$ ]. If this light falls on a silver plate having a work function of 4.7 eV , what will be the maximum kinetic energy of the photo electrons?
Select the correct option:
A
8.52 eV
B
7.72 eV
C
12.5 eV
D
6.82 eV
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Maximum Angular Frequency $=6.28 \times 10^7 \times 3 \times 10^8 \mathrm{rad} / \mathrm{s}$ $$ \begin{aligned} & \Rightarrow f_{\max }=3 \times 10^{15} \mathrm{~Hz} \\ & E_{\max }=h f_{\operatorname{nax}}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{15}}{1.6 \times 10^{-19}} \mathrm{ev} \\ & =12.375 \mathrm{ev} \square 12.38 \mathrm{eV} \\ & \Rightarrow K E_{\max }=12.38-4.7 \square 7.7 \mathrm{eV} \end{aligned} $$
Question Tags
JEE Main
Physics
Hard
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