Question

The magnetic field at the centre due to motion of electron in first Bohr orbit is B . The magnetic field due to motion of electron in second Bohr orbit at the centre will be (1) B/4 (2) B/8 (3) B/32 (4) B/64

Answer 1

Answer on Question #75045, Physics / Electromagnetism

Question. The magnetic field at the centre due to motion of electron in first Bohr orbit is $B$. The magnetic field due to motion of electron in second Bohr orbit at the centre will be (1) $B/4$; (2) $B/8$; (3) $B/32$; (4) $B/64$.

Solution.

$$B = \frac {\mu_ {0}}{4 \pi} \cdot \frac {e v \sin 9 0 {}^ {\circ}}{r ^ {2}} = \frac {\mu_ {0}}{4 \pi} \cdot \frac {e v}{r ^ {2}}$$

But, for Bohr orbits the quantum condition is

$$m v r = \frac {n h}{2 \pi} \quad \text{or} \quad v = \frac {n h}{2 \pi m r}$$ $$B = \frac {\mu_ {0}}{4 \pi} \cdot \frac {e n h}{2 \pi m r ^ {3}},$$

where $r = 53\cdot 10^{-12}n^2$

$$B = \frac {\mu_ {0}}{4 \pi} \cdot \frac {e n h}{2 \pi m n ^ {6} (5 3 \cdot 1 0 ^ {- 1 2}) ^ {3}}.$$

So,

$$B \propto \frac {1}{n ^ {5}}$$ $$\frac {B}{B _ {2}} = \frac {1}{1 ^ {5}} \cdot \frac {1}{2 ^ {5}} \rightarrow \frac {B}{B _ {2}} = 3 2 \rightarrow B _ {2} = \frac {B}{3 2}$$

Answer. (3) $B / 32$

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