The atomic mass of { }_{6} C^{12} is 12.00000 | vedclass

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(C) The nuclear reaction for removing a neutron from ${ }_{6} C^{13}$ is given by: ${ }_{6} C^{13} + 	ext{Energy} ightarrow { }_{6} C^{12} + { }_{0

Vedclass · Accepted Answer

$4.95$

Vedclass · Answer

(C) The nuclear reaction for removing a neutron from ${ }_{6} C^{13}$ is given by: ${ }_{6} C^{13} + 	ext{Energy} ightarrow { }_{6} C^{12} + { }_{0} n^{1}$.The mass defect $\Delta m$ is calculated as the difference between the mass of the products and the mass of the reactant:$\Delta m = (M({ }_{6} C^{12}) + M({ }_{0} n^{1})) - M({ }_{6} C^{13})$$\Delta m = (12.000000 + 1.008665) - 13.003354$$\Delta m = 13.008665 - 13.003354 = 0.005311 \ u$.The energy required is given by $E = \Delta m 	imes 931.5 \ MeV/u$:$E = 0.005311 	imes 931.5 \ MeV \approx 4.947 \ MeV \approx 4.95 \ MeV$.

The atomic mass of ${ }_{6} C^{12}$ is $12.000000 \ u$ and that of ${ }_{6} C^{13}$ is $13.003354 \ u$. The required energy to remove a neutron from ${ }_{6} C^{13}$,if the mass of a neutron is $1.008665 \ u$,will be: (in $MeV$)

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