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Question: In hydrogen atom, the electron is making \(6.6 \times 10 ^ { 15 } \mathrm { rev } / \mathrm { sec }...

In hydrogen atom, the electron is making 6.6×1015rev/sec6.6 \times 10 ^ { 15 } \mathrm { rev } / \mathrm { sec } around the nucleus in an orbit of radius 0.528 Å. The magnetic moment (A−m2)\left( A - m ^ { 2 } \right) will be

A

1×10−151 \times 10 ^ { - 15 }

B

1×10−101 \times 10 ^ { - 10 }

C

1×10−231 \times 10 ^ { - 23 }

D

1×10−271 \times 10 ^ { - 27 }

Answer

1×10−231 \times 10 ^ { - 23 }

Explanation

Solution

i=6.6×1015×1.6×10−19=10.5×10−4ampi = 6.6 \times 10 ^ { 15 } \times 1.6 \times 10 ^ { - 19 } = 10.5 \times 10 ^ { - 4 } \mathrm { amp }

A=πR2=3.142×(0.528)2×10−20 m2A = \pi R ^ { 2 } = 3.142 \times ( 0.528 ) ^ { 2 } \times 10 ^ { - 20 } \mathrm {~m} ^ { 2 }

⇒M=iA=10.5×10−4×3.142×(0.528)2×10−20M = i A = 10.5 \times 10 ^ { - 4 } \times 3.142 \times ( 0.528 ) ^ { 2 } \times 10 ^ { - 20 }

=10×10−24= 10 \times 10 ^ { - 24 } units =1×10−23= 1 \times 10 ^ { - 23 } units