Explanation
In hydrogen atom the speed of electron is $$v=\dfrac{{{e}^{2}}}{2{{\varepsilon }_{0}}nh}........(1)$$
Where, e is the charge of electron
n = 1 as first orbit
Dividing equation (1) by c
$$ \dfrac{v}{c}=\dfrac{{{e}^{2}}}{2{{\varepsilon }_{0}}ch} $$
$$ \dfrac{v}{c}=\dfrac{{{(1.6\times {{10}^{-19}})}^{2}}}{2\times 8.85\times {{10}^{-12}}\times 3\times {{10}^{8}}\times 6.6\times {{10}^{-34}}} $$
$$ \dfrac{v}{c}=\dfrac{1}{137} $$
Ratio is 1 : 137
In a cubical vessel $$1m \times 1m\times 1m$$ the gas molecules of diameter $$1.7 \times {10^{ - 8}}\;{\text{cm}}$$ are at a temperature 300 K and a pressure of $${10^{ - 4}}\;{\text{mm}}$$ mercury. The mean free path of the gas molecule is
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