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Physics Question on Current electricity

For a copper block, find the electric field which can give on an average 1 eV1 \,eV energy to a conduction electron. (The mean free path of conduction electrons in copper is given as 4×10−8 m4 \times 10^{-8}\, m )

A

2.62×107 V m−12.62 \times 10^{7}\,V\,m^{-1}

B

2.64×107 V m−12.64 \times 10^{7}\,V\,m^{-1}

C

2.5×107 V m−12.5 \times 10^{7}\,V\,m^{-1}

D

2.58×107 V m−12.58 \times 10^{7}\,V\,m^{-1}

Answer

2.5×107 V m−12.5 \times 10^{7}\,V\,m^{-1}

Explanation

Solution

Energy transfer to conduction electron
U=eV=e×E×dU=eV=e\times E \times d
[∵V=E×d][\because V=E \times d]
E=Ue×dE=\frac{U}{e\times d}
=1 eVe×4×10−8=\frac{1\,eV}{e\times4\times10^{-8}}
=1.6×10−19J1.6×10−19×4×10−8=\frac{1.6\times10^{-19}J}{1.6\times10^{-19}\times4\times10^{-8}}
=2.5×107V m−1=2.5\times10^{7}V\,m^{-1}