Question

The electric field maintained inside a conductor is 10N/C and they drift with a speed of $5\times {10}^{-4}m/s$
What is the relaxation time for the electrons?

• A. $\frac{5m}{e}\times {10}^{-5}$ seconds where m is the mass of the electrons and e is the charge
• B. $\frac{5e}{m}\times {10}^{-5}$ seconds where m is the mass of the electrons and e is the charge
• C. $\frac{m}{e}\times {10}^{-4}$ seconds where m is the mass of the electrons and e is the charge
• D. $\frac{5m}{e}\times {10}^{-4}$ seconds where m is the mass of the electrons and e is the charge

Answer 1

The correct option is A

$\frac{5m}{e}\times {10}^{-5}$ seconds where m is the mass of the electrons and e is the charge

From the videos you would have seen that,
$V_d=\frac{eE\tau }{m}$, where $\tau$ =Average time between successive collisions also called relaxation time.
Now you just need to plug in the values.
$5\times {10}^{-4}=10e\tau /m$
$\tau =\frac{5m}{e}\times {10}^{-5}seconds$.
You can use dimensional analysis to eliminate some options. Sometimes this method can be very useful.