Question

The electron mobility in a conductor is \(32~\text{cm}^{2}/\text{Vs}\). What is the relaxation time of free electrons? (Given: \(e=1.6\times10^{-19}~C,\; m=9.1\times10^{-31}~kg\))

• A. \(1.82\times10^{-12}\;s\)
• B. \(2.73\times10^{-12}\;s\)
• C. \(1.82\times10^{-14}\;s\)
• D. \(2.73\times10^{-14}\;s\)

Hint:

For questions involving electron mobility and relaxation time, always remember the relation \[ \boxed{\mu=\frac{e\tau}{m}} \] and convert mobility from \(\text{cm}^{2}/\text{Vs}\) to \(\text{m}^{2}/\text{Vs}\) before substitution. Useful conversion: \[ 1~\text{cm}^{2}=10^{-4}~\text{m}^{2}. \]

Answer 1

The Correct Option is A

Concept: Electron mobility (\(\mu\)) is defined as the drift velocity acquired by an electron per unit electric field. Mathematically, \[ \mu=\frac{v_d}{E} \] According to the free electron theory, \[ v_d=\frac{eE\tau}{m}, \] where \[ e=\text{Charge of electron}, \] \[ m=\text{Mass of electron}, \] \[ \tau=\text{Relaxation time}. \] Substituting the expression of drift velocity into the mobility equation, \[ \mu=\frac{e\tau}{m}. \] Hence, \[ \boxed{\tau=\frac{\mu m}{e}} \] This relation is frequently used to determine the relaxation time of free electrons.

Step 1: Convert the given mobility into SI units.
Given, \[ \mu=32~\text{cm}^{2}/\text{Vs}. \] Since, \[ 1~\text{cm}^{2}=10^{-4}\text{ m}^{2}, \] therefore, \[ 32~\text{cm}^{2} =32\times10^{-4}\text{ m}^{2} =3.2\times10^{-3}\text{ m}^{2}. \] Hence, \[ \boxed{\mu=3.2\times10^{-3}\text{ m}^{2}/\text{Vs}}. \]

Step 2: Write the formula for relaxation time.
The relation between mobility and relaxation time is \[ \mu=\frac{e\tau}{m}. \] Rearranging, \[ \tau=\frac{\mu m}{e}. \]

Step 3: Substitute the given values.
Given, \[ \mu=3.2\times10^{-3}\text{ m}^{2}/\text{Vs}, \] \[ m=9.1\times10^{-31}\text{ kg}, \] \[ e=1.6\times10^{-19}\text{ C}. \] Therefore, \[ \tau = \frac{(3.2\times10^{-3})(9.1\times10^{-31})} {1.6\times10^{-19}}. \] First, multiply the numerator, \[ 3.2\times9.1 =29.12, \] thus, \[ \tau = \frac{29.12\times10^{-34}} {1.6\times10^{-19}}. \] Now divide, \[ \frac{29.12}{1.6} =18.2. \] Hence, \[ \tau = 18.2\times10^{-15} = 1.82\times10^{-14}\;s. \] Therefore, \[ \boxed{\tau=1.82\times10^{-14}\;s.} \] Thus, the correct answer is \[ \boxed{\textbf{Option (C)}}. \]

Important Observation: The value obtained using the given data is \[ \boxed{1.82\times10^{-14}\;s.} \] \[ \boxed{\textbf{Option (C)}}. \]