A potential barrier of 0.4 V exists across a p-n junction. An electron enters the junction from the n- side with a speed of 6.0 × 105 ms ⁻¹. The speed with which electron enters the p side will be x/3 × 10⁵ ms ⁻¹ the value of x is _____________. (Given mass of electron = 9 × 10 - 31 kg, charge on electron = 1.6 × 10 - 19 C.) Numerical answer type. Work it out, then open the answer.

Answer: 14. Moving from n-side to p-side, the electron loses kinetic energy eV at the barrier. Initial KE = 1/2(9 × 10⁻³¹)(6 × 10⁵)² = 1.62 × 10⁻¹⁹ J. Loss = eV = 1.6 × 10⁻¹⁹ × 0.4 = 0.64 × 10⁻¹⁹ J, so final KE = 0.98 × 10⁻¹⁹ J. v² = 2 × 0.98 × 10⁻¹⁹9 × 10⁻³¹ = 2.18 × 10¹¹ ⇒ v = 4.67 × 10⁵ = 14/3 × 10⁵ m/s. So x = 14.

A potential barrier of 0.4 V exists across a p-n junction. An… | JEE Main 2022 (29 Jun, Shift 2)

Q68JEE Main 2022 · 29 JUN · SHIFT 2Difficult

P-N Junction Diode

A potential barrier of 0.4 V exists across a p-n junction. An electron enters the junction from the n- side with a speed of 6.0 $\times$ 105 ms ${}^{-1}$ . The speed with which electron enters the p side will be $\frac{x}{3} \times 10^{5}$ ms ${}^{-1}$ the value of x is _____________.

(Given mass of electron = 9 $\times$ 10 $-$ 31 kg, charge on electron = 1.6 $\times$ 10 $-$ 19 C.)

Numerical answer type. Work it out, then open the answer.

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Answer: 14

Moving from n-side to p-side, the electron loses kinetic energy $eV$ at the barrier.
Initial KE $= \frac{1}{2}(9 \times 10^{-31})(6 \times 10^{5})^2 = 1.62 \times 10^{-19}$ J.
Loss $= eV = 1.6 \times 10^{-19} \times 0.4 = 0.64 \times 10^{-19}$ J, so final KE $= 0.98 \times 10^{-19}$ J.
$v^2 = \frac{2 \times 0.98 \times 10^{-19}}{9 \times 10^{-31}} = 2.18 \times 10^{11} \Rightarrow v = 4.67 \times 10^5 = \frac{14}{3} \times 10^5$ m/s.
So $x = 14$ .

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