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Electric Field inside a Conductor under Electrodynamic Conditions

74) The nucle...

Question

74) The nucleus Fe⁵⁷ emits a y - ray of energy 14.4 keV. If mass of the nucleus is 56.935 amu, calculate the recoil energy of the nucleus. Take 1 amu = $1.66 \times 10^{- 27} k g .$

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Solution

$R e c o i l e n e r g y = \frac{14 . 4^{2}}{1862 * 56.935} = 2 e V R e g a r d s$

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