KCET · Physics · Dual Nature of Matter
Maximum velocity of the photoelectron emitted by a metal is \(1.8 \times 10^{6} \mathrm{~ms}^{-1}\). Take the value of specific charge of the electron is \(1.8 \times 10^{11} \mathrm{C} \mathrm{kg}^{-1}\). Then the stopping potential in volt is
- A 1
- B 3
- C 9
- D 6
Answer & Solution
Correct Answer
(C) 9
Step-by-step Solution
Detailed explanation
Given, \(v=1.8 \times 10^{6} \mathrm{~m} / \mathrm{s} ; \frac{e}{m}=1.8 \times 10^{11} \mathrm{C} / \mathrm{kg}\) We have, \(\quad e V_{0}=\frac{1}{2} m v^{2}\)
\(\begin{array}{ll}
\Rightarrow & V_{0} \frac{e}{m}=\frac{v^{2}}{2} \\
\Rightarrow \quad & V_{0} \times 1.8 \times 10^{11}=\frac{1.8 \times 1.8 \times\left(10^{6}\right)^{2}}{2} \\
\Rightarrow \quad & V_{0}=9 \mathrm{~V}
\end{array}\)
\(\begin{array}{ll}
\Rightarrow & V_{0} \frac{e}{m}=\frac{v^{2}}{2} \\
\Rightarrow \quad & V_{0} \times 1.8 \times 10^{11}=\frac{1.8 \times 1.8 \times\left(10^{6}\right)^{2}}{2} \\
\Rightarrow \quad & V_{0}=9 \mathrm{~V}
\end{array}\)
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