In hydrogen atom, the electron is making 5×1015 revolutions per second. If the radius of the orbit is 0.8×10−10m, then the magnetic field produced at the centre of the orbit is

B=μ0fe2rB=4π×10-7×5×1015×1.6×10-192×0.8×10-10B=62.8×10-1TB=6.28 T

In hydrogen atom, the electron is making 5×1015 revolutions per second. If the radius of the orbit is 0.8×10−10m, then the magnetic field produced at the centre of the orbit is

B=μ0fe2rB=4π×10-7×5×1015×1.6×10-192×0.8×10-10B=62.8×10-1TB=6.28 T

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In hydrogen atom, the electron is making $5 \times 10^{15}$ revolutions per second. If the radius of the orbit is $0.8 \times 10^{- 10} m$ , then the magnetic field produced at the centre of the orbit is

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1.57 T

3.14 T

4.71 T

6.28 T

answer is D.

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Detailed Solution

$B = \frac{μ_{0} f e}{2 r}$

$B = \frac{4 π \times 10^{- 7} \times 5 \times 10^{15} \times 1.6 \times 10^{- 19}}{2 \times 0.8 \times 10^{- 10}}$

$B = 62.8 \times 10^{- 1} T$

$B = 6.28 T$

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In hydrogen atom, the electron is making 5×1015 revolutions per second. If the radius of the orbit is 0.8×10−10m, then the magnetic field produced at the centre of the orbit is

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In hydrogen atom, the electron is making $5 \times 10^{15}$ revolutions per second. If the radius of the orbit is $0.8 \times 10^{- 10} m$ , then the magnetic field produced at the centre of the orbit is