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Electron Bohr orbit acceleration & energy radiation

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a) What is the acceleration of an electron in a circular Bohr orbit of radius
[4(pi)(epsilon_naught)(hbar^2)/(mass_electron)(e^2^)] in a hydrogen atom?

b) According to the classical electrodynamics, what is the rate at which the electron loses energy by radiation? Hint: the classical formula for energy lost by electromagnetic radiation is dE/dt=[(-e^2)(a^2)/(6)(pi)(epsilon_naught)(c^3)]

c) If the electron continues to radiate at this rate, how long does it take to reduce its enenergy from -13.6 eV to -2 x 13.6 eV?

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Solution Summary

Three parts of a problems related to electrons revolution in their orbits around a nucleus are solved. These include acceleration of the electron in the orbit, energy radiation rate, and energy loss due to radiation.

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a.
radius,
r = 4*pi * e_o * h_^2 / (m_e * e^2)

e_o = epsilon_naught
h_ = hbar
m_e = mass_electron

Because,
centripetal force = electric force:
m_e*v^2/r = e^2/(4*pi*e_o*r^2)

acceleration,
a = v^2/r = e^2/(4 * pi * e_o * r^2* m_e)
=> a = e^2/(4 * pi * e_o * (4*pi * e_o * h_^2 / (m_e * e^2))^2* m_e)
=> a = e^4 * m_e^2 /(4 * pi * e_o * 16*pi^2 * e_o^2 * h_^4 * m_e)
=> a = e^4 * m_e /(4 ...

Solution provided by:
Amrit Lal Ahuja, PhD (IP)
Education
  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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