Explanation

According to given data

We know that,

$$mvr=\dfrac{nh}{2\pi }$$

Or,

$$qvB=\dfrac{m{{v}^{2}}}{r}$$

So we find,

$$qB=\dfrac{mv}{r}$$

$$qB\left( \dfrac{nh}{2\pi mv} \right)=mv$$

$$ \left( \dfrac{1}{2}m{{v}^{2}} \right)=\dfrac{nhqB}{4\pi m} $$

$$ E=n\left( \dfrac{hqB}{4\pi m} \right) $$

This is the required solution.

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