In a p-n junction, an electric field of 5×10^5 V m exists in the depletion region. The minimum kinetic energy of a conduction electron, in order to…

Question

In a p-n junction, an electric field of 5×10^5 V m exists in the depletion region. The minimum kinetic energy of a conduction electron, in order to diffuse from n-side to the p-side, is found to be 3.2×10^(-20) J. The width of the depletion region is,

MARKS Content Team · Accepted Answer

Correct answer: (D) 4×10^(-5) cm. The kinetic energy in terms of electric potential, K.E.=eV, it means the value of electric potential, V=K.E./e=(3.2×10^(-20))/(1.6×10^(-19)), now the relation between electric field strength and electric potential in terms of the width of the depletion region, d=V/E=(3.2×10^(-20)1.6×10^(-19))/(5×10^5)=4×10^(-5) cm.

In a $p - n$ junction, an electric field of $5 \times 10^{5} V m$ exists in the depletion region. The minimum kinetic energy of a conduction electron, in order to diffuse from $n$ -side to the $p$ -side, is found to be $3.2 \times 10^{- 20} J$ . The width of the depletion region is,

Solution

Relevant Questions

In a p-n junction, an electric field of 5×105 V m exists in the depletion region. The minimum kinetic energy of a conduction electron, in order to diffuse from n-side to the p-side, is found to be 3.2×10-20 J. The width of the depletion region is,

Solution

Relevant Questions

In a $p - n$ junction, an electric field of $5 \times 10^{5} V m$ exists in the depletion region. The minimum kinetic energy of a conduction electron, in order to diffuse from $n$ -side to the $p$ -side, is found to be $3.2 \times 10^{- 20} J$ . The width of the depletion region is,