\(\frac{n h}{2 \pi}=m v r\) ..... \((ii)\)
Multiplying equation \((i)\) and \((ii)\),
\(\frac{q B n h}{2 \pi}=m^{2} v^{2}\)
Now multiplying both sides by \(\frac{1}{2 m}\)
\(n \frac{q B h}{4 \pi m}=\frac{1}{2} m v^{2}\)
i.e. \(\quad \mathrm{KE}=n\left[\frac{q B h}{4 \pi m}\right]\)
$\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{SI}\right.$ એકમ $\alpha$ કણનું દળ $=$ $\left.6.72 \times 10^{-27} \mathrm{~kg}\right)$