To calculate the mean free time ([tex]\(\tau\)[/tex]) given the mobility ([tex]\(\mu\)[/tex]) of an electron, the charge of an electron ([tex]\(e\)[/tex]), and the mass of an electron ([tex]\(m_e\)[/tex]), we can use the following relationship between these quantities:
[tex]\[ \mu = \frac{e \times \tau}{m_e} \][/tex]
We need to rearrange this formula to solve for [tex]\(\tau\)[/tex]:
[tex]\[ \tau = \frac{\mu \times m_e}{e} \][/tex]
Now let's substitute the given values into the formula:
- [tex]\(\mu = 3 \times 10^{-3} \, \text{m}^2 \, \text{V}^{-1} \, \text{s}^{-1}\)[/tex]
- [tex]\(e = 1.6 \times 10^{-19} \, \text{C}\)[/tex]
- [tex]\(m_e = 9.1 \times 10^{-31} \, \text{kg}\)[/tex]
Inserting these values:
[tex]\[ \tau = \frac{(3 \times 10^{-3} \, \text{m}^2 / \text{V} \cdot \text{s}) \times (9.1 \times 10^{-31} \, \text{kg})}{1.6 \times 10^{-19} \, \text{C}} \][/tex]
Carrying out the multiplication in the numerator:
[tex]\[ 3 \times 10^{-3} \times 9.1 \times 10^{-31} = 27.3 \times 10^{-34} \, \text{m}^2 \, \text{kg} \, \text{V}^{-1} \, \text{s}^{-1} \][/tex]
Now, divide by the charge ([tex]\(e\)[/tex]):
[tex]\[ \tau = \frac{27.3 \times 10^{-34}}{1.6 \times 10^{-19}} \, \text{s} \][/tex]
Performing the division:
[tex]\[ \tau = 1.70625 \times 10^{-14} \, \text{s} \][/tex]
Therefore, the mean free time [tex]\(\tau\)[/tex] is:
[tex]\[ \boxed{1.70625 \times 10^{-14} \, \text{s}} \][/tex]
