Answer:
The magnetic moment of an electron revolving in a circular orbit is given by the formula:
\mu = \frac{e \cdot \text{angular momentum}}{2m_e}μ=
2m
e
e⋅angular momentum
where:
\muμ is the magnetic moment,
ee is the charge of the electron,
m_em
is the mass of the electron.
The angular momentum of an electron in a circular orbit is quantized and given by:
\text{angular momentum} = \frac{nh}{2\pi}angular momentum=
2π
nh
nn is the principal quantum number of the orbit,
hh is the Planck's constant.
The charge of an electron is e = 1.6 \times 10^{-19}e=1.6×10
−19
Coulombs, the mass of an electron is m_e = 9.109 \times 10^{-31}m
=9.109×10
−31
kg, and Planck's constant is h = 6.626 \times 10^{-34}h=6.626×10
−34
Js.
Given the magnetic moment \mu = 5.024 \times 10^{-24}μ=5.024×10
−24
A-m² and the radius r = 2.2r=2.2 Å (which is 2.2 \times 10^{-10}2.2×10
−10
meters), we can use the formula for the magnetic moment to solve for the angular momentum.
First, the magnetic moment formula can be rearranged to find the angular momentum:
\text{angular momentum} = \frac{2m_e \cdot \mu}{e}angular momentum=
⋅μ
\text{angular momentum} = \frac{2 \times 9.109 \times 10^{-31} \times 5.024 \times 10^{-24}}{1.6 \times 10^{-19}}angular momentum=
1.6×10
2×9.109×10
×5.024×10
\text{angular momentum} \approx 1.43 \times 10^{-34} \, \text{kg m²/s}angular momentum≈1.43×10
kg m²/s
Now, the angular momentum for a circular orbit is quantized:
Setting these equations equal to each other:
\frac{nh}{2\pi} = 1.43 \times 10^{-34} \, \text{kg m²/s}
=1.43×10
Solving for the frequency ff (frequency of revolution):
f = \frac{1}{T} = \frac{v}{2 \pi r} = \frac{v}{2 \pi \cdot 2.2 \times 10^{-10}}f=
T
1
=
2πr
v
2π⋅2.2×10
f = \frac{\frac{nh}{2\pi}}{2 \pi r} = \frac{1.43 \times 10^{-34}}{2 \pi \cdot 2.2 \times 10^{-10}}f=
1.43×10
Calculating the frequency will give the number of revolutions per second.
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Answers & Comments
Answer:
The magnetic moment of an electron revolving in a circular orbit is given by the formula:
\mu = \frac{e \cdot \text{angular momentum}}{2m_e}μ=
2m
e
e⋅angular momentum
where:
\muμ is the magnetic moment,
ee is the charge of the electron,
m_em
e
is the mass of the electron.
The angular momentum of an electron in a circular orbit is quantized and given by:
\text{angular momentum} = \frac{nh}{2\pi}angular momentum=
2π
nh
where:
nn is the principal quantum number of the orbit,
hh is the Planck's constant.
The charge of an electron is e = 1.6 \times 10^{-19}e=1.6×10
−19
Coulombs, the mass of an electron is m_e = 9.109 \times 10^{-31}m
e
=9.109×10
−31
kg, and Planck's constant is h = 6.626 \times 10^{-34}h=6.626×10
−34
Js.
Given the magnetic moment \mu = 5.024 \times 10^{-24}μ=5.024×10
−24
A-m² and the radius r = 2.2r=2.2 Å (which is 2.2 \times 10^{-10}2.2×10
−10
meters), we can use the formula for the magnetic moment to solve for the angular momentum.
First, the magnetic moment formula can be rearranged to find the angular momentum:
\text{angular momentum} = \frac{2m_e \cdot \mu}{e}angular momentum=
e
2m
e
⋅μ
\text{angular momentum} = \frac{2 \times 9.109 \times 10^{-31} \times 5.024 \times 10^{-24}}{1.6 \times 10^{-19}}angular momentum=
1.6×10
−19
2×9.109×10
−31
×5.024×10
−24
\text{angular momentum} \approx 1.43 \times 10^{-34} \, \text{kg m²/s}angular momentum≈1.43×10
−34
kg m²/s
Now, the angular momentum for a circular orbit is quantized:
\text{angular momentum} = \frac{nh}{2\pi}angular momentum=
2π
nh
Setting these equations equal to each other:
\frac{nh}{2\pi} = 1.43 \times 10^{-34} \, \text{kg m²/s}
2π
nh
=1.43×10
−34
kg m²/s
Solving for the frequency ff (frequency of revolution):
f = \frac{1}{T} = \frac{v}{2 \pi r} = \frac{v}{2 \pi \cdot 2.2 \times 10^{-10}}f=
T
1
=
2πr
v
=
2π⋅2.2×10
−10
v
f = \frac{\frac{nh}{2\pi}}{2 \pi r} = \frac{1.43 \times 10^{-34}}{2 \pi \cdot 2.2 \times 10^{-10}}f=
2πr
2π
nh
=
2π⋅2.2×10
−10
1.43×10
−34
Calculating the frequency will give the number of revolutions per second.