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An electron revolving in an orbit of radius 0.5Angstroms in a hydrogen atom executes 10^16 revolutions per second.The magnetic moment of electron due to its orbital motion will be. 1)1.256×10^-23 Am^2 2)653×10^-26 Am^2. 3)10^-3 Am^2. 4)256×10^-26 Am^2?
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An electron revolving in an orbit of radius 0.5Angstroms in a hydrogen...
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An electron revolving in an orbit of radius 0.5Angstroms in a hydrogen...
Given:
Radius of the orbit, r = 0.5 Å = 0.5 × 10^(-10) m
Revolutions per second, n = 10^16
To find:
The magnetic moment of the electron due to its orbital motion.
Solution:
The magnetic moment of a charged particle due to its orbital motion is given by the formula:
μ = e * A * n
where
μ is the magnetic moment,
e is the charge of the electron (1.6 × 10^(-19) C),
A is the area of the orbit, and
n is the number of revolutions per second.
Calculating the Area of the Orbit:
The area of a circle is given by the formula:
A = π * r^2
where
A is the area of the circle, and
r is the radius of the circle.
Substituting the given values:
A = π * (0.5 × 10^(-10))^2
A = π * (0.25 × 10^(-20))
A = (π/4) × 10^(-20) m^2
Calculating the Magnetic Moment:
Substituting the values of e, A, and n:
μ = (1.6 × 10^(-19) C) * [(π/4) × 10^(-20) m^2] * (10^16 rev/s)
μ = (1.6 × 10^(-19) C) * (π/4) * (10^(-20) m^2) * (10^16 rev/s)
μ = (1.6 × π/4) * (10^(-19 - 20 + 16)) C m^2
μ = (0.4π) * (10^(-23)) C m^2
Calculating the numerical value:
μ ≈ 1.256 × 10^(-23) C m^2
Therefore, the magnetic moment of the electron due to its orbital motion is approximately 1.256 × 10^(-23) C m^2.
Answer:
The correct option is 1) 1.256 × 10^(-23) Am^2.
Radius of the orbit, r = 0.5 Å = 0.5 × 10^(-10) m
Revolutions per second, n = 10^16
To find:
The magnetic moment of the electron due to its orbital motion.
Solution:
The magnetic moment of a charged particle due to its orbital motion is given by the formula:
μ = e * A * n
where
μ is the magnetic moment,
e is the charge of the electron (1.6 × 10^(-19) C),
A is the area of the orbit, and
n is the number of revolutions per second.
Calculating the Area of the Orbit:
The area of a circle is given by the formula:
A = π * r^2
where
A is the area of the circle, and
r is the radius of the circle.
Substituting the given values:
A = π * (0.5 × 10^(-10))^2
A = π * (0.25 × 10^(-20))
A = (π/4) × 10^(-20) m^2
Calculating the Magnetic Moment:
Substituting the values of e, A, and n:
μ = (1.6 × 10^(-19) C) * [(π/4) × 10^(-20) m^2] * (10^16 rev/s)
μ = (1.6 × 10^(-19) C) * (π/4) * (10^(-20) m^2) * (10^16 rev/s)
μ = (1.6 × π/4) * (10^(-19 - 20 + 16)) C m^2
μ = (0.4π) * (10^(-23)) C m^2
Calculating the numerical value:
μ ≈ 1.256 × 10^(-23) C m^2
Therefore, the magnetic moment of the electron due to its orbital motion is approximately 1.256 × 10^(-23) C m^2.
Answer:
The correct option is 1) 1.256 × 10^(-23) Am^2.
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An electron revolving in an orbit of radius 0.5Angstroms in a hydrogen atom executes 10^16 revolutions per second.The magnetic moment of electron due to its orbital motion will be. 1)1.256×10^-23 Am^2 2)653×10^-26 Am^2. 3)10^-3 Am^2. 4)256×10^-26 Am^2?
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