JEE Main 2026 — Physics Modern Physics

2026mcqeasy

In a semiconductor p-n diode, the doping concentrations on p-side and n-side are $10^{15}\text{ atoms/cm}^3$ and $10^{18}\text{ atoms/cm}^3$ , respectively. Which one of the following statements is true ?

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Options

A
Widths of depletion region on either side of the interface are equal
B
The depletion region width is more on p-side compared to that in n-side
C
The depletion region width is more on n-side compared to that in p-side
D
No depletion region forms because of unequal doping concentrations on p and n-sides

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Solution

In a p-n junction diode, the total charge on both sides of the depletion region must be equal to maintain charge neutrality.

Let $x_p$ and $x_n$ be the widths of the depletion region on the p-side and n-side, respectively. Let $N_A$ and $N_D$ be the doping concentrations on the p-side and n-side.

The charge neutrality condition is given by:

$N_A x_p = N_D x_n$

Given $N_A = 10^{15} \text{ atoms/cm}^3$ and $N_D = 10^{18} \text{ atoms/cm}^3$ .

Substituting the values, we get:

$10^{15} x_p = 10^{18} x_n$

$x_p = 10^3 x_n$

Since $x_p > x_n$ , the depletion region penetrates more into the lightly doped side. Therefore, the depletion region width is more on the p-side compared to that in the n-side.

Answer: The depletion region width is more on p-side compared to that in n-side

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Options

Widths of depletion region on either side of the interface are equal

The depletion region width is more on p-side compared to that in n-side

The depletion region width is more on n-side compared to that in p-side

No depletion region forms because of unequal doping concentrations on p and n-sides

Solution

In a p-n junction diode, the total charge on both sides of the depletion region must be equal to maintain charge neutrality.

Let $x_p$ and $x_n$ be the widths of the depletion region on the p-side and n-side, respectively. Let $N_A$ and $N_D$ be the doping concentrations on the p-side and n-side.

The charge neutrality condition is given by:

$N_A x_p = N_D x_n$

Given $N_A = 10^{15} \text{ atoms/cm}^3$ and $N_D = 10^{18} \text{ atoms/cm}^3$ .

Substituting the values, we get:

$10^{15} x_p = 10^{18} x_n$

$x_p = 10^3 x_n$

Since $x_p > x_n$ , the depletion region penetrates more into the lightly doped side. Therefore, the depletion region width is more on the p-side compared to that in the n-side.

Answer: The depletion region width is more on p-side compared to that in n-side

JEE Main 2026 — Physics Modern Physics

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JEE Main 2026 — Physics Modern Physics

Options

Did you get this right?

Solution

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JEE Main Modern Physics in other years

More JEE Main Modern Physics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Modern Physics