CBSE 2026 · Region 4 · Set 1 · Q21 · 2 marks
The hole concentration in an intrinsic semiconductor is $\displaystyle 5 \times 10^{8} \mathrm{~m}^{-3}$. When it is doped with certain impurity, the electron concentration becomes $\displaystyle 4 \times 10^{12} \mathrm{~m}^{-3}$. Find the new value of the hole concentration. Also identify the type of new semiconductor formed after doping.
Marking-scheme solution
$\displaystyle n_{e} n_{h}=n_{i}^{2}$
Hole concentration $\displaystyle \left(n_{h}\right)=\dfrac{n_{i}^{2}}{n_{e}}=\dfrac{5 \times 10^{8} \times 5 \times 10^{8}}{4 \times 10^{12}}$
$\displaystyle n_{h}=6.25 \times 10^{4} \mathrm{~m}^{-3}$
The new semiconductor formed is an n-type semiconductor.
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CBSE Class 12 Physics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.