Numerical Problems Based on Mobility of Charge Carriers for Class 12 Physics

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Numerical Problems on Mobility of Charge Carriers

Class 12 Physics · Current Electricity
SQ

Mobility and Conductivity Practice Set

Apply $\mu = \frac{v_d}{E}$ and $\sigma = ne\mu$
Ch 3 · Current Electricity
1
A potential difference of 4.5 V is applied across a conductor of length 0.1 m. If the drift velocity of electrons is $1.5 \times 10^{-4}\text{ ms}^{-1}$, find the electron mobility.
Answer $3.33 \times 10^{-6}\text{ m}^2\text{V}^{-1}\text{s}^{-1}$ 📝
Detailed Solution

Potential difference $V = 4.5\text{ V}$, length $l = 0.1\text{ m}$, drift velocity $v_d = 1.5 \times 10^{-4}\text{ ms}^{-1}$.

First, calculate the electric field ($E$) across the conductor:

$$E = \frac{V}{l} = \frac{4.5}{0.1} = 45\text{ Vm}^{-1}$$

Now, compute the electron mobility ($\mu$) using the relation $\mu = \frac{v_d}{E}$:

$$\mu = \frac{1.5 \times 10^{-4}}{45} = \frac{1}{30} \times 10^{-4}$$
$$\mu = 0.0333 \times 10^{-4} = 3.33 \times 10^{-6}\text{ m}^2\text{V}^{-1}\text{s}^{-1}$$
2
The number density of electrons in copper is $8.5 \times 10^{28}\text{ m}^{-3}$. A current of 1A flows through a copper wire of length 0.24 m and area of cross-section $1.2\text{ mm}^2$, when connected to a battery of 3 V. Find the electron mobility.
Answer $4.9 \times 10^{-6}\text{ m}^2\text{V}^{-1}\text{s}^{-1}$ 📝
Detailed Solution

Given $n = 8.5 \times 10^{28}\text{ m}^{-3}$, Current $I = 1\text{ A}$, Length $l = 0.24\text{ m}$, Area $A = 1.2\text{ mm}^2 = 1.2 \times 10^{-6}\text{ m}^2$, Potential difference $V = 3\text{ V}$, and charge of electron $e = 1.6 \times 10^{-19}\text{ C}$.

1. Calculate the drift velocity ($v_d$):

$$v_d = \frac{I}{nAe} = \frac{1}{(8.5 \times 10^{28}) \times (1.2 \times 10^{-6}) \times (1.6 \times 10^{-19})}$$
$$v_d = \frac{1}{16.32 \times 10^3} = 6.127 \times 10^{-5}\text{ ms}^{-1}$$

2. Calculate the electric field ($E$):

$$E = \frac{V}{l} = \frac{3}{0.24} = 12.5\text{ Vm}^{-1}$$

3. Calculate electron mobility ($\mu$):

$$\mu = \frac{v_d}{E} = \frac{6.127 \times 10^{-5}}{12.5} = 4.9 \times 10^{-6}\text{ m}^2\text{V}^{-1}\text{s}^{-1}$$
3
Mobilities of electrons and holes in a sample of intrinsic germanium at room temperature are $0.54\text{ m}^2\text{V}^{-1}\text{s}^{-1}$ and $0.18\text{ m}^2\text{V}^{-1}\text{s}^{-1}$ respectively. If the electron and hole densities are equal to $3.6 \times 10^{19}\text{ m}^{-3}$, calculate the germanium conductivity.
Answer $4.147\text{ Sm}^{-1}$ 📝
Detailed Solution

Electron mobility $\mu_e = 0.54\text{ m}^2\text{V}^{-1}\text{s}^{-1}$, Hole mobility $\mu_h = 0.18\text{ m}^2\text{V}^{-1}\text{s}^{-1}$.
Carrier density $n_i = n_e = n_h = 3.6 \times 10^{19}\text{ m}^{-3}$.
Elementary charge $e = 1.6 \times 10^{-19}\text{ C}$.

The total conductivity ($\sigma$) of an intrinsic semiconductor is the sum of the conductivities due to electrons and holes:

$$\sigma = n_e e \mu_e + n_h e \mu_h = n_i e (\mu_e + \mu_h)$$

Substitute the given values:

$$\sigma = (3.6 \times 10^{19}) \times (1.6 \times 10^{-19}) \times (0.54 + 0.18)$$
$$\sigma = (3.6 \times 1.6) \times (0.72)$$
$$\sigma = 5.76 \times 0.72 = 4.1472\text{ Sm}^{-1}$$

Rounding to three decimal places:

$$\sigma \approx 4.147\text{ Sm}^{-1}$$

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