The mean free path of conduction electrons in copper is about 4 × 10-8 m . Find the electric field which can give, on an average, 2 eV energy to a conduction electron in a block of copper.

Question

The mean free path of conduction electrons in copper is about 4 × 10-8 m . Find the electric field which can give, on an average, 2 eV energy to a conduction electron in a block of copper. (A) 5 × 107 V / m (B) 5 × 106 V / m (C) 5 × 10-6 V / m (D) 5 × 10-7 V / m

Tardigrade Admin · Accepted Answer

Mean free path d=4 × 10-8 m , E =? Energy of electron =2 eV F = e E Work done on electron when it moves through distance d = eED. This work done is equal to the energy transfered to the electron, so ∴ eEd =2 eV Ed =2 V E=(2 V/d)=(2/4 × 10-8) =5 × 107 V / m

Q. The mean free path of conduction electrons in copper is about $4 \times 1 0^{- 8} m .$ Find the electric field which can give, on an average, $2 e V$ energy to a conduction electron in a block of copper.

$5 \times 1 0^{7} V / m$

$5 \times 1 0^{6} V / m$

$5 \times 1 0^{- 6} V / m$

$5 \times 1 0^{- 7} V / m$

Q. The mean free path of conduction electrons in copper is about 4×10−8m. Find the electric field which can give, on an average, 2eV energy to a conduction electron in a block of copper.

5×107V/m

5×106V/m

5×10−6V/m

5×10−7V/m

Mean free path d=4×10−8m,E=? Energy of electron =2eV F=eE Work done on electron when it moves through distance d=eED. This work done is equal to the energy transfered to the electron, so ∴eEd=2eV Ed=2V E=d2V​=4×10−82​ =5×107V/m

Q. The mean free path of conduction electrons in copper is about $4 \times 1 0^{- 8} m .$ Find the electric field which can give, on an average, $2 e V$ energy to a conduction electron in a block of copper.

$5 \times 1 0^{7} V / m$

$5 \times 1 0^{6} V / m$

$5 \times 1 0^{- 6} V / m$

$5 \times 1 0^{- 7} V / m$