A nucleus 23N10 undergoes decay and becomes 23Na11. Calculate the maximum kinetic energy of electron emitting assuming that the daughter nucleus and anti- neutrino carry negligible kinetic energy.Mass of 23Ne10 = 22.994466u. Mass of 23Na11 = 22.989770.1u = 931.5 MeV/c^2

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Solution

$D e a r S t u d e n t, m a s s d i f f e r e n c e = 22.994466 - 22.989770 = 0.004696 e n e r g y q u i v a l e n t t o m a s s d i f f e r e n c e = 0.004696 * 931.5 M e v = 4.374324 M e v . t h i s e n e r g y i s t h e K E o f t h e e l e c t r o n K E o f e l e c t r o n = 4.374324 M e v R e g a r d s$

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_{11}^{23} N e

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