JEE PHYSICS MCQ: SEMI CONDUCTOR DEVICES

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Application +4 | -1

Calculate the current (in A) produced in a germanium plate of thickness

0.25 \mathrm{~mm}

and area

1 \mathrm{~cm}^2

when a potential difference of

1.5 \mathrm{~V}

is applied across the thickness. The hole density in germanium is

2 \times 10^{19} \mathrm{~m}^{-3}

and the mobilities of electrons and holes are

350 \mathrm{~cm}^2 \mathrm{~V}^{-1} \mathrm{~s}^{-1}

and

1200 \mathrm{~cm}^2 V^{-1} \mathrm{~s}^{-1}

respectively.

View answer and explanation

Correct option: A

0.902 \mathrm{~A} \text {. }

Option A (

0.902 \mathrm{~A} \text {. }

) is correct because it is the calculated current using the given hole density, mobilities, and applied potential difference.

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