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Question: An electron revolving in an orbit of radius \(0.5 A\) in a hydrogen atom executes \(10^{16}\) revolu...

An electron revolving in an orbit of radius 0.5A0.5 A in a hydrogen atom executes 101610^{16} revolutions per second. The magnetic moment of electron due to its orbital motion will be:

Explanation

Solution

Number of electrons and the radius of the orbit are given. Charge is equal to the number of electrons multiplied by the charge on an electron. Current is equal to the ratio of charge to the total time taken. Magnetic moment is equal to the current times of the area of orbit in which electrons are revolving.

Complete step-by-step solution:
Number of electrons, n=1016n = 10^{16}
Radius of orbit, r=0.5A=5×10−11mr = 0.5 A = 5 \times 10^{-11} m
Charge on an electron, e=1.6×10−19Ce = 1.6 \times 10^{-19} C
Total Charge, Q=ne=1016×1.6×10−19Q = ne =10^{16} \times 1.6 \times 10^{-19}
Q=1.6×10−3CQ = 1.6 \times 10^{-3} C
Current, I=QTI = \dfrac{Q}{T}
Time taken, T=1second T = 1 second
Now, Value of current, I=QT=1.6×10−3C1sI = \dfrac{Q}{T} =\dfrac{1.6 \times 10^{-3} C}{1 s}
I=1.6×10−3AI= 1.6 \times 10^{-3} A
Magnetic moment M=IA=1.6×10−3×πr2M = IA = 1. 6 \times 10^{-3} \times \pi r^{2}
  ⟹  M=IA=1.6×10−3×π(5×10−11)2\implies M = IA = 1. 6 \times 10^{-3} \times \pi \left( 5 \times 10^{-11} \right)^{2}
  ⟹  M=IA=1.6×10−3×3.14(5×10−11)2\implies M = IA = 1. 6 \times 10^{-3} \times 3.14 \left( 5 \times 10^{-11} \right)^{2}
  ⟹  M=1.256×10−23Am−2\implies M = 1.256 \times 10^{-23} Am^{-2}
The magnetic moment of an electron due to its orbital motion will be 1.256×10−23Am−21.256 \times 10^{-23} Am^{-2}.

Note: The magnetic moment is equal to the magnetic force and orientation that generates a magnetic field. Examples: Objects that produce magnetic moments involve circuits of electric current, strong magnets, elementary particles, different molecules, and many celestial objects. The magnetic moment is a measurement of its trend to get organized through a magnetic field.