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JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter
The magnetic field associated with a light wave is given, at the origin, by \(B = B_0 [sin\,(3.14 \times 10^7)\,ct\,+\,sin\,(6.28 \times 10^7)\,ct]\) If this light falls on a silver plate having a work function of \(4.7\,eV,\) what will be the maximum kinetic energy of the photo electrons? .............. \(eV\)
- A \(6.82\)
- B \(12.5\)
- C \(8.52\)
- D \(7.72\)
Answer & Solution
Correct Answer
(D) \(7.72\)
Step-by-step Solution
Detailed explanation
\(\mathrm{KE}_{\max } =\mathrm{hv}_{\max }-\phi\) \( = \frac{{(6.6 \times {{10}^{ - 34}})(6.28 \times {{10}^7})(3 \times {{10}^8})}}{{1.6 \times {{10}^{ - 19}} \times 2 \times 3.14}} - 4.7\) \(=12.37-4.7=7.67 \mathrm{eV}\)
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