COMEDK · Physics · 27. Dual Nature of Matter
The photoelectric current of voltage in a certain experiment is \(1.5 \mathrm{~V}\). What is the maximum kinetic energy of photoelectrons emitted?
- A \(24 \times 10^{-19} \mathrm{~J}\)
- B \(-24 \times 10^{-19} \mathrm{~J}\)
- C \(2.4 \times 10^{-19} \mathrm{~J}\)
- D \(2.04 \times 10^{-19} \mathrm{~J}\)
Answer & Solution
Correct Answer
(C) \(2.4 \times 10^{-19} \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
Given, \(V_{0}=15 \mathrm{~V}\) Maximum kinetic energy of emitted photoelectron, \(\begin{aligned} K_{\max } &=e V_{0} \\ &=1.6 \times 10^{-19} \times 1.5=2.4 \times 10^{-19} \mathrm{~J} \end{aligned}\)
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