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======= ####21. The mean free path of electrons in a metal is $4 \times 10^{- 8} m$ . The electric field which can give on an average $2 e V$ energy to an electron in the metal will be in units of $V / m$

c3eec34ec6b1fad69db54a20ad4b2dca40c2aa54 (a) $5 \times 10^{- 11}$

(b) $8 \times 10^{- 11}$

(d) $8 \times 10^{7}$

[2009]

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Answer:

Correct Answer: 21. (c)

Solution:

$\Rightarrow e V_{o} = 2 e V \Rightarrow V_{o} = 2$ volt

$\begin{aligned} E & = \frac{V_{O}}{d} = \frac{2}{4 \times 10^{- 8}} \\ = 0.5 \times 10^{8} = 5 \times 10^{7} V m^{- 1} \end{aligned}$

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Electric Charges and Fields - Result Question 21

21. 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 of V/m

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Answer:

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21. The mean free path of electrons in a metal is $4 \times 10^{- 8} m$ . The electric field which can give on an average $2 e V$ energy to an electron in the metal will be in units of $V / m$