Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the appl

Q: Why are we ignoring the holes in this calculation?

In an $n$-type semiconductor, electrons are the majority charge carriers, and their concentration is vastly greater than that of holes. The question also explicitly states to ignore the hole contribution to simplify the calculation.

Q: What is the value of $e$ and why isn't it given in the question?

The symbol $e$ represents the elementary charge of an electron, which is a fundamental physical constant equal to $1.6 \times 10^{-19} \text{ C}$. In JEE exams, you are expected to have standard constants like this memorized.

Q: How do I quickly calculate $\frac{1}{2.56}$ without a calculator?

You can use approximation! Notice that $2.56$ is very close to $2.5$. Since $2.5 = \frac{10}{4}$, its reciprocal is $\frac{4}{10} = 0.4$. This perfectly matches option (c).

Q: What is the difference between drift velocity and mobility?

Drift velocity ($\vec{v}_d$) is the actual average speed at which electrons move under an electric field. Mobility ($\mu_e$) is a property of the material that defines how much drift velocity is achieved *per unit* of applied electric field ($\mu_e = \frac{v_d}{E}$).

Q: Would the formula change if it was a $p$-type semiconductor?

Yes! For a $p$-type semiconductor, holes are the majority carriers. If we ignore electrons, the conductivity formula would be $\sigma = n_h e \mu_h$, where $n_h$ is the hole density and $\mu_h$ is the hole mobility.

Question

Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If for an $n$-type semiconductor, the density of electrons is $10^{19} 	ext{ m}^{-3}$ and their mobility is $1.6 	ext{ m}^2 	ext{ (V-s)}^{-1}$, then the resistivity of the semiconductor (since, it is an $n$-type semiconductor contribution of holes is ignored) is close to

Options:

(A) $2 \ \Omega	ext{-m}$

(B) $0.2 \ \Omega	ext{-m}$

(C) $0.4 \ \Omega	ext{-m}$

(D) $4 \ \Omega	ext{-m}$

Sigma Percentile · Accepted Answer

### The Physics of Drifting Electrons

Imagine a block of an $n$-type semiconductor. When we apply an external electric field $\vec{E}$ across it, the free electrons inside don't just sit there—they experience an electrostatic force and begin to drift. Because electrons carry a negative charge, they drift in the direction opposite to the applied electric field with a certain drift velocity $\vec{v}_d$.

The ease with which these electrons move through the semiconductor lattice is quantified by a property called **mobility** ($\mu_e$). The question provides us with the electron density $n_e = 10^{19} 	ext{ m}^{-3}$ and their mobility $\mu_e = 1.6 	ext{ m}^2	ext{V}^{-1}	ext{s}^{-1}$. Since it's an $n$-type semiconductor, the majority charge carriers are electrons, and we are explicitly told to ignore the contribution of holes.

### The Conductivity Equation

The ability of a material to conduct electric current is measured by its **conductivity** ($\sigma$). For a semiconductor where only electrons are contributing to the current, the conductivity is directly proportional to the number density of electrons, the elementary charge $e$, and the electron mobility.

The master equation for conductivity is:

$$ \sigma = n_e e \mu_e $$

This equation makes intuitive sense: more electrons ($n_e$), a larger charge per carrier ($e$), or a higher ability to move ($\mu_e$) will all result in a higher overall conductivity.

### Substituting the Values

Now, let's plug the given values into our conductivity equation. We know the elementary charge $e = 1.6 	imes 10^{-19} 	ext{ C}$.

$$ \sigma = (10^{19}) 	imes (1.6 	imes 10^{-19}) 	imes (1.6) $$

Here comes the beautiful part of the calculation. The examiner has set up the numbers so that the powers of 10 perfectly cancel each other out. The $10^{19}$ from the electron density neutralizes the $10^{-19}$ from the elementary charge.

$$ \sigma = 1.6 	imes 1.6 = 2.56 \ \Omega^{-1}	ext{m}^{-1} $$

### The Reciprocal Relationship: Resistivity

We have successfully found the conductivity, but the question asks for the **resistivity** ($ho$). Resistivity is simply the reciprocal of conductivity. It measures how strongly a material opposes the flow of electric current.

$$ ho = \frac{1}{\sigma} $$

Substituting our calculated value of $\sigma$:

$$ ho = \frac{1}{2.56} $$

To calculate this quickly, you can approximate $2.56$ as roughly $2.5$, which is $\frac{10}{4}$. The reciprocal would be $\frac{4}{10} = 0.4$. If we calculate it precisely, $\frac{1}{2.56} \approx 0.3906 \ \Omega	ext{-m}$.

Looking at the options, $0.39 \ \Omega	ext{-m}$ is extremely close to **$0.4 \ \Omega	ext{-m}$**. Therefore, the correct choice is option (c).

Select Answer:

Visualized Solution

Visualizing the Semiconductor

Conductivity Formula

Substituting Values

Calculating Conductivity

Resistivity Formula

Final Calculation

Conclusion

The Sigma Insight: Energy Bands in Solids and Semiconductors

The Physics of Drifting Electrons

The Conductivity Equation

Substituting the Values

The Reciprocal Relationship

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