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JEE MAIN 2019
09-04-2019 - S1
Question
The electric field of light wave is given as $\vec{E}=10^{-3} \cos \left(\frac{2 \pi x}{5 \times 10^{-7}}-2 \pi \times 6 \times 10^{14} t\right) \hat{x} \frac{N}{C}$. This light falls on a metal plate of work function 2eV. The stopping potential of the photoelectrons is :
Given, E (in eV) = $\frac{12375}{\lambda(\text { in } Å)}$
Select the correct option:
A
0.48 V
B
2.48 V
C
0.72 V
D
2.0 V
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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Question Tags
JEE Main
Physics
Easy
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