JEE Main 2022 — Physics Modern Physics

2022integerhard

A potential barrier of $0.4V$ exists across a p-n junction. An electron enters the junction from the $n$ -side with a speed of $6.0\times {10}^{5}{ms}^{-1}$ . The speed with which electron enters the $p$ side will be $\frac{x}{3}\times {10}^{5}{ms}^{-1}$ , then the value of $x$ is _____.: (Given mass of electron $=9\times {10}^{-31}\mathrm{kg}$ , charge on electron $=1.6\times {10}^{-19}C$ .)

Official previous-year question

Held on 29 Jun 2022 · Verified 6 Jul 2026.

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Solution

JEE Main 2022 Physics, Semiconductors — solution figure

Applying conservation of energy

$-e{V}_{n}+\frac{1}{2}m{v}_{i}^{2}=-e{V}_{p}+\frac{1}{2}m{v}_{f}^{2}$

We know, ${V}_{n}-{V}_{p}=$ potential barrier $=0.4V$

Then, we have, $-e({V}_{n}-{V}_{p})+\frac{1}{2}m{v}_{i}^{2}=\frac{1}{2}m{v}_{f}^{2}$

$\Rightarrow -1.6\times {10}^{-19}\times 0.4+\frac{1}{2}\times 9\times {10}^{-31}\times 36\times {10}^{10}=\frac{1}{2}\times 9\times {10}^{-31}{v}_{f}^{2}$

$\Rightarrow -0.64\times {10}^{-19}+1.62\times {10}^{-19}=\frac{1}{2}\times 9\times {10}^{-31}{v}_{f}^{2}$

$\Rightarrow 0.98\times {10}^{-19}=4.5\times {10}^{-31}{v}_{f}^{2}$

$\Rightarrow {v}_{f}^{2}=\frac{0.98\times {10}^{12}}{4.5}$

$\Rightarrow {v}_{f}=0.466666\times {10}^{6}$

$=4.6666\times {10}^{5}m{s}^{-1}=\frac{14}{3}\times {10}^{5}m{s}^{-1}$

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JEE Main 2022 — Physics Modern Physics

2022integerhard

Official previous-year question

Held on 29 Jun 2022 · Verified 6 Jul 2026.

Did you get this right?

Solution

JEE Main 2022 Physics, Semiconductors — solution figure

Applying conservation of energy

$-e{V}_{n}+\frac{1}{2}m{v}_{i}^{2}=-e{V}_{p}+\frac{1}{2}m{v}_{f}^{2}$

We know, ${V}_{n}-{V}_{p}=$ potential barrier $=0.4V$

Then, we have, $-e({V}_{n}-{V}_{p})+\frac{1}{2}m{v}_{i}^{2}=\frac{1}{2}m{v}_{f}^{2}$

$\Rightarrow -1.6\times {10}^{-19}\times 0.4+\frac{1}{2}\times 9\times {10}^{-31}\times 36\times {10}^{10}=\frac{1}{2}\times 9\times {10}^{-31}{v}_{f}^{2}$

$\Rightarrow -0.64\times {10}^{-19}+1.62\times {10}^{-19}=\frac{1}{2}\times 9\times {10}^{-31}{v}_{f}^{2}$

$\Rightarrow 0.98\times {10}^{-19}=4.5\times {10}^{-31}{v}_{f}^{2}$

$\Rightarrow {v}_{f}^{2}=\frac{0.98\times {10}^{12}}{4.5}$

$\Rightarrow {v}_{f}=0.466666\times {10}^{6}$

$=4.6666\times {10}^{5}m{s}^{-1}=\frac{14}{3}\times {10}^{5}m{s}^{-1}$

Your note

JEE Main Modern Physics in other years

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JEE Main 2022 — Physics Modern Physics

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Solution

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JEE Main Modern Physics in other years

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JEE Main 2022 — Physics Modern Physics

Did you get this right?

Solution

Your note

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