KCET · Physics · Dual Nature of Matter
A hot filament liberates an electron with zero initial velocity. The anode potential is \(1200 \mathrm{~V}\). The speed of the electron when it strikes the anode is
- A \(1.5 \times 10^{5} \mathrm{~ms}^{-1}\)
- B \(2.5 \times 10^{6} \mathrm{~ms}^{-1}\)
- C \(2.1 \times 10^{7} \mathrm{~ms}^{-1}\)
- D \(2.5 \times 10^{8} \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(C) \(2.1 \times 10^{7} \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Given, anode potential, \(V=1200 \mathrm{~V}\) Electron will accelerate with the effect of anode potential.
Hence, \(\frac{1}{2} m v^{2}=\mathrm{eV}\)
\(\begin{aligned}
v &=\sqrt{\frac{2 e V}{m}}=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 1200}{9.1 \times 10^{-31}}} \\
&=\sqrt{421.98 \times 10^{12}}=20.5 \times 10^{6} \mathrm{~ms}^{-1} \\
&=2.05 \times 10^{7} \mathrm{~ms}^{-1} \approx 2.1 \times 10^{7} \mathrm{~ms}^{-1}
\end{aligned}\)
Hence, \(\frac{1}{2} m v^{2}=\mathrm{eV}\)
\(\begin{aligned}
v &=\sqrt{\frac{2 e V}{m}}=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 1200}{9.1 \times 10^{-31}}} \\
&=\sqrt{421.98 \times 10^{12}}=20.5 \times 10^{6} \mathrm{~ms}^{-1} \\
&=2.05 \times 10^{7} \mathrm{~ms}^{-1} \approx 2.1 \times 10^{7} \mathrm{~ms}^{-1}
\end{aligned}\)
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