The electric field of light wave is given as
\(\vec{E}=10^{-3}\cos\left(\frac{2\pi x}{5\times10^{-7}}-2\pi\times6\times10^{14}t\right)\hat{x}\ \frac{\mathrm{N}}{C}\)
This light falls on a metal plate of work function 2 eV. The stopping potential of the photoelectrons is :
Given, \(E\) (in eV) \(=12375/\lambda\) (in \(\mathrm{\AA}\))
\(\vec{E}=10^{-3}\cos\left(\frac{2\pi x}{5\times10^{-7}}-2\pi\times6\times10^{14}t\right)\hat{x}\ \frac{\mathrm{N}}{C}\)
This light falls on a metal plate of work function 2 eV. The stopping potential of the photoelectrons is :
Given, \(E\) (in eV) \(=12375/\lambda\) (in \(\mathrm{\AA}\))
Show Hint
- 2.48 V
- 0.48 V
- 0.72 V
- 2.0 V
The Correct Option is B
Solution and Explanation
Photon energy \(E=\frac{12375}{5000}=2.475\) eV.
Einstein's equation: \(eV_0=E-\phi=2.475-2=0.475\) eV, i.e. \(V_0\approx0.48\) V.
Answer: (2) 0.48 V
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