Correct Answer - Option 2 : Mobility increase with increase in mass of charge
CONCEPT:
- Relaxation time:(τ)
- Relaxation time is the time gap between two successive electron collisions in a conductor.
- Drift Velocity:(Vd)
Subatomic particles like electrons move in random directions all the time. When electrons are subjected to an electric field they do move randomly, but they slowly drift in one direction, in the direction of the electric field applied. The net velocity at which these electrons drift is known as drift velocity.
- The average velocity attained by charged particles, (eg. electrons) in a material due to an electric field.
- The SI unit of drift velocity is m/s.
- The relationship between the relaxation time (τ) and drift velocity (Vd) is given below:
⇒\(V_{d} = \frac{eEτ}{m}\)
- Where, Vd = drift velocity, e = charge of electron, E = Electric field, m = mass of electron ,τ = Relaxation time
- So the expression for relaxation time (τ ) is,
⇒ \(\tau = \frac{mV_{d}}{eE}\)
- Mobility(µ): The mobility of a charge carrier is defined as the drift velocity of the charge carrier per unit electric field.
⇒\(\mu = \frac{V_{d}}{E}\)
EXPLAINATION:
⇒\(\mu = \frac{V_{d}}{E}..... 1\)
⇒\(V_{d} = \frac{eE\tau }{m}.....2\)
Putting equation 2 in 1
⇒\(\mu = \frac{e\tau }{m}\)
Therefore,
⇒\(\mu \propto \tau \)
⇒\(\mu \propto V_{d} \)
⇒\(\mu \propto \frac{1}{m} \)
⇒\(\mu \propto \frac{1}{E}\)
- On increasing drift velocity and relaxation time, mobility increases
- On decreasing mass and electric field, mobility increases. Hence option (2) is correct.
