TS EAMCET · Physics · Current Electricity
Find the mobility of electron in a wire, if its average collision time is \(9.1 \times 10^{-5} \mathrm{~s}\). (charge of electron \(=1.6 \times 10^{-19} \mathrm{C}\) and mass of electron \(\left.=9.1 \times 10^{-31} \mathrm{~kg}\right)\)
- A \(9.1 \times 10^{-3} \mathrm{~m}^2 / \mathrm{V}-\mathrm{s}\)
- B \(1.6 \times 10^{-3} \mathrm{~m}^2 / \mathrm{V}-\mathrm{s}\)
- C \(1.75 \times 10^{-3} \mathrm{~m}^2 / \mathrm{V}-\mathrm{s}\)
- D \(1 \times 10^{-3} \mathrm{~m}^2 / \mathrm{V}-\mathrm{s}\)
Answer & Solution
Correct Answer
(B) \(1.6 \times 10^{-3} \mathrm{~m}^2 / \mathrm{V}-\mathrm{s}\)
Step-by-step Solution
Detailed explanation
Mobility is given as \[ \begin{aligned} & \mu=\frac{V_d}{E}=\frac{e \tau}{m}=\frac{1.6 \times 10^{-19} \times 9.1 \times 10^{-15}}{9.1 \times 10^{-31}} \\ & =1.6 \times 10^{-3} \frac{\mathrm{m}^2}{V-S} \end{aligned} \]
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