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JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2019
- Hard
- MCQ
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 ................ \(V\)
Pick an option to check your answer.
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
\(\mathrm{KE}_{\max }=\mathrm{E}-\phi\)
\(\mathrm{eV}_{\mathrm{s}}=\mathrm{E}-\phi\)
\(\mathrm{eV}_{\mathrm{s}}=2.475-2\)
\(\mathrm{eV}_{\mathrm{o}}=0.475\, \mathrm{eV}\)
\(\mathrm{V}_{\mathrm{o}}=0.48\, \mathrm{V}\)
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