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.)

Solution

Physics sir · 4 steps

The barrier slows the electron down, so subtract eV from its kinetic energy and find the new speed.

1. Direction of force. The barrier field points from n to p, so it pushes an electron back towards the n-side. Crossing it, the electron loses kinetic energy equal to eV.

   1/2mv² = 1/2mu² - eV ⇒ v² = u² - 2eV/m

2. Put in numbers. u² = (6 × 10⁵)² = 36 × 10¹⁰ = 324/9 × 10¹⁰ m² s⁻². Also 2eV/m = (2 × 1.6 × 10⁻¹⁹ × 0.4)/(9 × 10⁻³¹) = 128/9 × 10¹⁰ m² s⁻².
3. Find the speed. v² = (324 - 128)/9 × 10¹⁰ = 196/9 × 10¹⁰, so v = 14/3 × 10⁵ m s⁻¹ ≈ 4.67 × 10⁵ m s⁻¹.
4. So. Matching v = x/3 × 10⁵ m s⁻¹ gives x = 14.

The idea to keep. A charge crossing a potential difference V gains or loses exactly qV of kinetic energy. Electrons going from n to p climb the barrier, so they slow down.

Answer from the NTA final answer key. Solution by PrepWiser.