Q137 / 180 · KCET 2016 · Physics
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
Then time of fall
- Correct answerA\( t=\sqrt{\frac{2 h m}{e E}} \)
- B\( t=\frac{2 h m}{e E} \)
- C\( t=\sqrt{\frac{2 e E}{h m}} \)
- D\( t=\frac{2 e E}{h m} \)
Solution
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
\[ \begin{array}{l} t^{2}=\frac{2 m h}{e E} \\ \Rightarrow t=\sqrt{\frac{2 m h}{e E}} \end{array} \]
Hence, the correct answer is option: 'A'.