In a p-n junction diode, an electric field of magnitude 2 × 105 V / m exists in the depletion region. A particle with charge -3 e can diffuse from n - side to p - side, if it has minimum kinetic energy 0.6 eV. The width of the depletion region of the p-n junction is

Question

In a p-n junction diode, an electric field of magnitude 2 × 105 V / m exists in the depletion region. A particle with charge -3 e can diffuse from n - side to p - side, if it has minimum kinetic energy 0.6 eV. The width of the depletion region of the p-n junction is (A) 300 nm (B) 600 nm (C) 1000 nm (D) 1200 nm

Tardigrade Admin · Accepted Answer

For depletion width d, developed potential is V=E ⋅ d=2 × 105 × d For charged particle, experienced potential is V=( text Energy / text Charge )=(0 ⋅ 6 eV /3 e)=0 ⋅ 2 V So, 2 × 105 × d=0 ⋅ 2 ⇒ d =10-6 m or 1000 nm

Q. In a $p - n$ junction diode, an electric field of magnitude $2 \times 1 0^{5} V / m$ exists in the depletion region. A particle with charge $- 3 e$ can diffuse from $n$ - side to $p$ - side, if it has minimum kinetic energy $0.6 e V$ .
The width of the depletion region of the $p - n$ junction is

300 nm

600 nm

1000 nm

1200 nm

For depletion width $d$ , developed potential is
$V = E \cdot d = 2 \times 1 0^{5} \times d$
For charged particle, experienced potential is
$V = \frac{Energy}{Charge} = \frac{0 \cdot 6 e V}{3 e} = 0 \cdot 2 V$
So, $2 \times 1 0^{5} \times d = 0 \cdot 2$
$\Rightarrow d = 1 0^{- 6} m$ or $1000 nm$

Q. In a p−n junction diode, an electric field of magnitude 2×105V/m exists in the depletion region. A particle with charge −3e can diffuse from n - side to p - side, if it has minimum kinetic energy 0.6eV. The width of the depletion region of the p−n junction is