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An electron of mass \( \mathrm{m} \), charge e falls through a distance \( \mathrm{h} \) meter in a uniform electric field \( \mathrm{E} \).
Then time of fall
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Then time of fall
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Verified Answer
The correct answer is:
\( t=\sqrt{\frac{2 h m}{e E}} \)
Given, mass of electron \( =m \); charge \( =e \); height \( =h \)
Using \( S+u t+\frac{1}{2} a t^{2} \), we have
\[
\begin{array}{l}
h=0+\frac{1}{2} \frac{e E}{m} t^{2} \\
\Rightarrow h=\frac{1}{2} \frac{e E}{m} t^{2}
\end{array}
\]
On rearranging, we get
\[
\begin{array}{l}
t^{2}=\frac{2 m h}{e E} \\
\Rightarrow t=\sqrt{\frac{2 m h}{e E}}
\end{array}
\]
Using \( S+u t+\frac{1}{2} a t^{2} \), we have
\[
\begin{array}{l}
h=0+\frac{1}{2} \frac{e E}{m} t^{2} \\
\Rightarrow h=\frac{1}{2} \frac{e E}{m} t^{2}
\end{array}
\]
On rearranging, we get
\[
\begin{array}{l}
t^{2}=\frac{2 m h}{e E} \\
\Rightarrow t=\sqrt{\frac{2 m h}{e E}}
\end{array}
\]
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