Physics Circle home Student Login

Home›Chapters›Current Electricity PYQs›Electric Current, Drift Velocity & Current Density›Question 33

NEET 2009 (AIPMT) · Electric Current, Drift Velocity & Current Density

Electric Current, Drift Velocity & Current Density question from NEET 2009 (AIPMT)

Q33NEET 2009 · AIPMTMedium

Electric Current, Drift Velocity & Current Density

−-8

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

Show answer and solution

Answer: (c)

  1. Energy gained over one mean free path: eEλ=2 eVeE\lambda = 2\text{ eV}
  2. E=2 Vλ=24×10−8E = \dfrac{2\text{ V}}{\lambda} = \dfrac{2}{4 \times 10^{-8}}
  3. E=5×107 V/mE = 5 \times 10^{7}\text{ V/m}
  4. So the answer is (c).
Report a mistake

More questions on Electric Current, Drift Velocity & Current Density