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Current Electricity question

2009 · Q133
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Current Electricity question

2009 · Q133

NEETPhysicsCurrent ElectricityMCQ+4 / −1
The mean free path of electrons in a metal is 4 ×\times× 10−-−8 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
  1. A
    5×10−115 \times {10^{ - 11}}5×10−11
  2. B
    8 ×\times× 10−-−11
  3. C
    5 ×\times× 107
  4. D
    8 ×\times× 107
View written solutionFree

Correct answer: C

Given, Energy = 2 eV = eEλ\lambda λ

∴\therefore∴ E = 2eVeλ=24×10−8=5×107{{2eV} \over {e\lambda }} = {2 \over {4 \times {{10}^{ - 8}}}} = 5 \times {10^7}eλ2eV​=4×10−82​=5×107 V/m

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