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JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter

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 \(2\,eV.\) The stopping potential of the photoelectrons is ................ \(V\)

  1. A \(0.48\)
  2. B \(2.48\)
  3. C \(0.72\)
  4. D \(2\)
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Answer & Solution

Correct Answer

(A) \(0.48\)

Step-by-step Solution

Detailed explanation

\(\omega=6 \times 10^{14} \times 2 \pi\) \(f=6 \times 10^{14}\) \(\lambda = \frac{C}{f} = \frac{{3 \times {{10}^8}}}{{6 \times {{10}^{14}}}} = 5000\,\,\mathop {\text{A}}\limits^o \) Energy of photon \(\Rightarrow \frac{12375}{5000}=2.475\, \mathrm{eV}\) From Einstein's equation…
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