Find the energy liberated in the beta decay of `._6^(14)C " to " ._7^(14)N` as represented by Eq.(iii). Equation (iii) refers to nuclei. Adding six electrons to both sides of Eq.(iii) gives ` .-6^(14)C "atom" rarr .-7^(14)N`.

Find the energy liberated in the beta decay of `._6^(14)C " to " ._7^(

1.	Find the energy liberated in the beta decay of `._6^(14)C " to " ._7^(14)N` as represented by Eq.(iii). Equation (iii) refers to nuclei. Adding six electrons to both sides of Eq.(iii) gives ` .-6^(14)C "atom" rarr .-7^(14)N`.
Answer» We know that `._(6)^(14)C` has a mass of `14.003 074u` . Here, the mass differrence between the initial and final states is `Delta m=14.003 242 u -14.003 74 u = 0.000 168 u` This corresponds to an energy release of `E = (0.000 168 u)(931 . 494 MeV//u)=0.156 MeV`.

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Find the energy liberated in the beta decay of `._6^(14)C " to " ._7^(14)N` as represented by Eq.(iii). Equation (iii) refers to nuclei. Adding six electrons to both sides of Eq.(iii) gives ` .-6^(14)C "atom" rarr .-7^(14)N`.

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