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When an electron of charge e and mass m moves with a velocity v about the nuclear charge Ze is circular orbit of radius r, the potential energy of the electrons is given by [1994]
- a)Ze2/r
- b)-Ze2/r
- c)Ze-2/r
- d)mv2/r
Correct answer is option 'B'. Can you explain this answer?
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When an electron of charge e and mass m moves with a velocity v about ...
P.E. = work done 
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When an electron of charge e and mass m moves with a velocity v about ...
Explanation:
The potential energy of an electron in a circular orbit of radius r around a nucleus of charge Ze can be calculated using the Coulomb's law of electrostatics. The Coulomb's law states that the force between two charges q1 and q2 separated by a distance r is given by:
F = (1/4πε0) * (q1q2/r^2)
where ε0 is the permittivity of free space.
When an electron moves in a circular orbit around a nucleus of charge Ze, the electrostatic force between the electron and the nucleus provides the necessary centripetal force to keep the electron in its orbit. The magnitude of the electrostatic force is given by:
F = (1/4πε0) * (eZe/r^2)
where e is the charge of the electron and m is its mass.
The centripetal force required to keep the electron in its circular orbit is given by:
F = mv^2/r
where v is the velocity of the electron.
Setting these two forces equal to each other, we get:
m v^2/r = (1/4πε0) * (eZe/r^2)
Multiplying both sides by r, we get:
mv^2 = (1/4πε0) * (eZe/r)
The potential energy of the electron in the circular orbit is given by the work done by the electrostatic force in bringing the electron from infinity to the orbit. This can be calculated using the formula:
U = -∫∞r F.dr
where U is the potential energy, F is the electrostatic force, and r is the distance between the electron and the nucleus.
Integrating the above expression, we get:
U = (-1/4πε0) * (eZe/r)
Therefore, the potential energy of the electron in a circular orbit of radius r around a nucleus of charge Ze is given by:
U = - (1/4πε0) * (eZe/r)
which is option B.
The potential energy of an electron in a circular orbit of radius r around a nucleus of charge Ze can be calculated using the Coulomb's law of electrostatics. The Coulomb's law states that the force between two charges q1 and q2 separated by a distance r is given by:
F = (1/4πε0) * (q1q2/r^2)
where ε0 is the permittivity of free space.
When an electron moves in a circular orbit around a nucleus of charge Ze, the electrostatic force between the electron and the nucleus provides the necessary centripetal force to keep the electron in its orbit. The magnitude of the electrostatic force is given by:
F = (1/4πε0) * (eZe/r^2)
where e is the charge of the electron and m is its mass.
The centripetal force required to keep the electron in its circular orbit is given by:
F = mv^2/r
where v is the velocity of the electron.
Setting these two forces equal to each other, we get:
m v^2/r = (1/4πε0) * (eZe/r^2)
Multiplying both sides by r, we get:
mv^2 = (1/4πε0) * (eZe/r)
The potential energy of the electron in the circular orbit is given by the work done by the electrostatic force in bringing the electron from infinity to the orbit. This can be calculated using the formula:
U = -∫∞r F.dr
where U is the potential energy, F is the electrostatic force, and r is the distance between the electron and the nucleus.
Integrating the above expression, we get:
U = (-1/4πε0) * (eZe/r)
Therefore, the potential energy of the electron in a circular orbit of radius r around a nucleus of charge Ze is given by:
U = - (1/4πε0) * (eZe/r)
which is option B.
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When an electron of charge e and mass m moves with a velocity v about the nuclear charge Ze is circular orbit of radius r, the potential energy of the electrons is given by [1994]a)Ze2/rb)-Ze2/rc)Ze-2/rd)mv2/rCorrect answer is option 'B'. Can you explain this answer?
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