The mean free path of electrons in a metal is 4 × 10-8m. The electric field which can give on an average 2 eV energy to an electron in the metal will be in units V/m

Question

The mean free path of electrons in a metal is 4 × 10-8m. The electric field which can give on an average 2 eV energy to an electron in the metal will be in units V/m (A) 5 × 10-11 (B) 8 × 10-11 (C) 5 × 107 (D) 8 × 107

Tardigrade Admin · Accepted Answer

Energy =2 eV =e E λ ∴ E=(2 e V/e λ)=(2/4 × 10-8)=5 × 107 V / m

Q. The mean free path of electrons in a metal is $4 \times 1 0^{- 8} m .$ The electric field which can give on an average $2 e V$ energy to an electron in the metal will be in units $V / m$

$5 \times 1 0^{- 11}$

$8 \times 1 0^{- 11}$

$5 \times 1 0^{7}$

$8 \times 1 0^{7}$

Energy $= 2 e V = e E λ$
$∴ E = \frac{2 e V}{e λ} = \frac{2}{4 \times 1 0 ^{- 8}} = 5 \times 1 0^{7} V / m$

Q. The mean free path of electrons in a metal is 4×10−8m. The electric field which can give on an average 2eV energy to an electron in the metal will be in units V/m

5×10−11

8×10−11

5×107

8×107