The electric field of light wave is given as

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 $2 \mathrm{eV}$. The stopping potential of the photo-electrons is:

Given, $\mathrm{E}($ in $\mathrm{eV})=\frac{12375}{\lambda(\text { in } A)}$

  1. (1) $2.0 \mathrm{~V}$

  2. (2) $0.72 \mathrm{~V}$

  3. (3) $0.48 \mathrm{~V}$

  4. (4) $2.48 \mathrm{~V}$


Correct Option: 3

Solution:

(3) Here $\omega=2 \pi \times 6 \times 10^{14}$

$\Rightarrow \mathrm{f}=6 \times 10^{14} \mathrm{~Hz}$

Wavelength

$\lambda=\frac{c}{f}=\frac{3 \times 10^{8}}{6 \times 10^{14}}=0.5 \times 10^{-6} \mathrm{~m}=5000 A

Given $E=\frac{12375}{5000}=2.48 \mathrm{eV}$

Using $\mathrm{E}=\mathrm{W}+\mathrm{eV}_{\mathrm{s}}$

$\Rightarrow 2.48=2+\mathrm{eV}_{\mathrm{s}}$

or $\mathrm{V}_{\mathrm{s}}=0.48 \mathrm{~V}$

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