Q. Thermal speed of free electrons is of the order of :

Rajasthan PMTRajasthan PMT 2004 Report Error

A

$ 10^{-5}\,m/s $

B

$ 10^{2}\,m/s $

C

$ 10^{8}\,m/s $

D

$ 10^{5}\,m/s $

Solution:

Thermal energy of free electron is
$=\frac{3}{2} k T$
Let $v$ be the thermal speed of electron, then
$\frac{1}{2} m v^{2}=\frac{3}{2} k T$
$v=\sqrt{\frac{3 k T}{m}}$
$\Rightarrow v=\sqrt{\left[\frac{3 \times 1.38 \times 10^{-23} \times 300}{9.1 \times $$10^{-31}}\right]}$
$=10^{5} \,m / s$

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