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)

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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