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Calculte the orbital magnetic dipole momeng of the electron in the second bohr orbit of hydrogen atom the radius of o is 2.126 A and orbital speed of electron is 1.09×10 -6 m/sec.?
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Calculte the orbital magnetic dipole momeng of the electron in the sec...
Calculation of Orbital Magnetic Dipole Moment of Electron in the Second Bohr Orbit of Hydrogen Atom
The orbital magnetic dipole moment of an electron in the second Bohr orbit of a hydrogen atom can be calculated using the following formula:
μ = (evr)/(2m)
Where,
e = charge of the electron = 1.602 x 10^-19 C
v = orbital speed of the electron = 1.09 x 10^-6 m/s
r = radius of the orbit = 2.126 A = 2.126 x 10^-10 m
m = mass of the electron = 9.109 x 10^-31 kg
Substituting the Values in the Formula
μ = (1.602 x 10^-19 C x 1.09 x 10^-6 m/s x 2.126 x 10^-10 m)/(2 x 9.109 x 10^-31 kg)
On solving the above equation, we get:
μ = 9.274 x 10^-24 A m^2
Therefore, the orbital magnetic dipole moment of the electron in the second Bohr orbit of a hydrogen atom is 9.274 x 10^-24 A m^2.
Explanation
The orbital magnetic dipole moment of an electron is a measure of the strength of its magnetic field due to its orbital motion around the nucleus. It is given by the product of the charge, speed, and radius of the electron's orbit, divided by twice the mass of the electron.
In this case, we have used the values of charge, speed, and radius of the electron in the second Bohr orbit of a hydrogen atom, and the mass of the electron. On substituting these values in the formula, we get the value of the orbital magnetic dipole moment of the electron in the given orbit.
Conclusion
The orbital magnetic dipole moment of the electron in the second Bohr orbit of a hydrogen atom is 9.274 x 10^-24 A m^2.
The orbital magnetic dipole moment of an electron in the second Bohr orbit of a hydrogen atom can be calculated using the following formula:
μ = (evr)/(2m)
Where,
e = charge of the electron = 1.602 x 10^-19 C
v = orbital speed of the electron = 1.09 x 10^-6 m/s
r = radius of the orbit = 2.126 A = 2.126 x 10^-10 m
m = mass of the electron = 9.109 x 10^-31 kg
Substituting the Values in the Formula
μ = (1.602 x 10^-19 C x 1.09 x 10^-6 m/s x 2.126 x 10^-10 m)/(2 x 9.109 x 10^-31 kg)
On solving the above equation, we get:
μ = 9.274 x 10^-24 A m^2
Therefore, the orbital magnetic dipole moment of the electron in the second Bohr orbit of a hydrogen atom is 9.274 x 10^-24 A m^2.
Explanation
The orbital magnetic dipole moment of an electron is a measure of the strength of its magnetic field due to its orbital motion around the nucleus. It is given by the product of the charge, speed, and radius of the electron's orbit, divided by twice the mass of the electron.
In this case, we have used the values of charge, speed, and radius of the electron in the second Bohr orbit of a hydrogen atom, and the mass of the electron. On substituting these values in the formula, we get the value of the orbital magnetic dipole moment of the electron in the given orbit.
Conclusion
The orbital magnetic dipole moment of the electron in the second Bohr orbit of a hydrogen atom is 9.274 x 10^-24 A m^2.
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Calculte the orbital magnetic dipole momeng of the electron in the sec...
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Calculte the orbital magnetic dipole momeng of the electron in the second bohr orbit of hydrogen atom the radius of o is 2.126 A and orbital speed of electron is 1.09×10 -6 m/sec.?
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Calculte the orbital magnetic dipole momeng of the electron in the second bohr orbit of hydrogen atom the radius of o is 2.126 A and orbital speed of electron is 1.09×10 -6 m/sec.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Calculte the orbital magnetic dipole momeng of the electron in the second bohr orbit of hydrogen atom the radius of o is 2.126 A and orbital speed of electron is 1.09×10 -6 m/sec.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculte the orbital magnetic dipole momeng of the electron in the second bohr orbit of hydrogen atom the radius of o is 2.126 A and orbital speed of electron is 1.09×10 -6 m/sec.?.
Calculte the orbital magnetic dipole momeng of the electron in the second bohr orbit of hydrogen atom the radius of o is 2.126 A and orbital speed of electron is 1.09×10 -6 m/sec.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Calculte the orbital magnetic dipole momeng of the electron in the second bohr orbit of hydrogen atom the radius of o is 2.126 A and orbital speed of electron is 1.09×10 -6 m/sec.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculte the orbital magnetic dipole momeng of the electron in the second bohr orbit of hydrogen atom the radius of o is 2.126 A and orbital speed of electron is 1.09×10 -6 m/sec.?.
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