- #1

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My solution was this: find the momentum of the electron (via de Broglie relation) and use a very standard kinetic energy formula, like this (assuming minimum energy state has a wavelength of twice the nucleus' diameter):

[tex] p=h/\lambda=h/2d [/tex]

[tex] E_K = p^2/2m_e = \frac{h^2}{8d^2m_e} [/tex]

The professors marked this as incorrect, and instead gave this solution ([tex] E_0 [/tex] is electron rest mass, 0.511 MeV):

[tex] p=h/\lambda=h/2d [/tex]

[tex] E_K = [(pc)^2 + E_0^2]^{1/2} - E_o [/tex]

These give drastically different results, but I'm just curious why the professor's approximation is physically more valid than my own.