COMEDK · Physics · 27. Dual Nature of Matter
When a beam of 10.6 eV photons of intensity \(2.0 \mathrm{~W} / \mathrm{m}^2\) falls on a platinum surface of area \(1.0 \times 10^{-4} \mathrm{~m}^2\) and work function 5.6 eV . \(0.53 \%\) of the incident photons eject photoelectrons. The number of photoelectrons emitted per second is
- A \(6.25 \times 10^{11}\)
- B \(4.25 \times 10^{10}\)
- C \(6.25 \times 10^{10}\)
- D \(4.25 \times 10^{11}\)
Answer & Solution
Correct Answer
(A) \(6.25 \times 10^{11}\)
Step-by-step Solution
Detailed explanation
Energy of incident photons, \(\begin{aligned} E_1 & =10.6 \mathrm{eV}=10.6 \times 1.6 \times 10^{-19} \mathrm{~J} \\ & =16.96 \times 10^{-19} \mathrm{~J} \end{aligned}\) Energy incident per unit area per unit time (intensity) \(=2 \mathrm{~J}\) \(\therefore\) Number of photons…
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