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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
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Verified Answer
The correct answer is:
$2.1 \times 10^{7} \mathrm{~ms}^{-1}$
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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