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Physics Question 45 – JEE-MAIN 2026

In a semiconductor p-n diode, the doping concentrations on p-side and n-side are 1015 atoms/cm3 and 1018 atoms/cm3, respectively. Which one of the following statements is true?

The depletion region forms due to the diffusion of charge carriers across the p-n junction, creating an electric field and uncovering immobile ionized impurity atoms.

Step 1: Relate Depletion Width to Doping Concentration✦ Active

In a p-n junction, the total charge on the p-side of the depletion region must be equal in magnitude to the total charge on the n-side to maintain charge neutrality. This implies that the product of doping concentration and depletion width on each side is equal:

NAWp=NDWn

where NA is the acceptor concentration on the p-side, ND is the donor concentration on the n-side, Wp is the depletion width on the p-side, and Wn is the depletion width on the n-side.

Step 2: Apply Given Doping Concentrations○ Expand

Given doping concentrations are NA=1015 atoms/cm3 (p-side) and ND=1018 atoms/cm3 (n-side). From the relationship derived in Step 1, we can find the ratio of the depletion widths:

WpWn=NDNA=10181015=103
💡 Teacher's Secret Hint

Remember that the depletion region extends more into the lightly doped side.

Step 3: Determine Relative Widths and Select Correct Statement○ Expand

The ratio WpWn=1000 indicates that Wp=1000×Wn. This means the depletion region width on the p-side (Wp) is significantly larger than that on the n-side (Wn). The p-side is lightly doped (1015), and the n-side is heavily doped (1018). Therefore, the depletion region extends more into the lightly doped p-side.

Comparing this conclusion with the given options:

Option 1: Incorrect, widths are unequal.

Option 2: Correct, the depletion region width is more on the p-side compared to that in the n-side.

Option 3: Incorrect.

Option 4: Incorrect, a depletion region always forms in a p-n junction regardless of doping concentration equality.

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