Since an electron on a circular orbit around a proton has a centripetal acceleration, it should radiate energy according to the Larmor relation dE dt 2 3 (q 2 40) (a 2 c 3 ) where q, a, 0 , and c are, respectively, the electron charge, its acceleration, the vacuum permittivity, and the velocity of ...

The Larmor formula relates the rate of energy loss of a charged particle in motion due to the emissi ...

Since an electron on a circular orbit around a proton has a centripetal acceleration, it should radiate

Question:

Since an electron on a circular orbit around a proton has a centripetal acceleration, it should radiate energy according to the Larmor relation dE/dt¼-2/3 (q²/4πε0) (a²/c³) where q, a, ε⁰, and c are, respectively, the electron charge, its acceleration, the vacuum permittivity, and the velocity of light in vacuum. Therefore, in classical mechanics, it should spiral and crash on the nucleus. How long would this decay take, supposing that the size of the initial orbit is 10 -10 m and the nucleus is a point charge (i.e., radius ¼ 0)?

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