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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 photo-electrons is :

Given, $E\left ( in\; eV \right )=\frac{12375}{\lambda \left ( in\; \AA \right )}$

Option 1)
$2.0 \: V$

Option 2)
$0.72\: V$

Option 3)
$0.48\: V$

Option 4)
$2.48\: V$

Answers (1)

best_answer

$E=10^{-3}cos\left ( \frac{2\pi x}{5\times 10^{-7}}-2\pi \times 6\times 10^{14}t \right )\hat x\frac{N}{C}$

$\phi = Work \; function=2eV$

$E=\frac{12375}{\lambda\; in \AA } \; ev$

$K=\frac{2\pi }{5\times10^{-7}}\; m\; \; \; \Rightarrow K=\frac{2\pi }{\lambda }$

$\lambda =5\times 10^{-7}m$

$E=\frac{12375}{5000}=2.475\; PV$

Stopping Potential= $\frac{\frac{hc}{\lambda }-\phi }{e}$

$=\frac{2.475\; ev-2\; ev}{e}$

$=0.475 \; V$

Option 1)

$2.0 \: V$

Option 2)

$0.72\: V$

Option 3)

$0.48\: V$

Option 4)

$2.48\: V$

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