The binding energy per nucleon for ^6C^{12} n | vedclass

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(B) The number of protons $Z = 6$ and the number of neutrons $N = 12 - 6 = 6$.The mass defect $\Delta m$ is given by:$\Delta m = [Z \cdot m_H + N \cdo

Vedclass · Accepted Answer

$7.675$

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(B) The number of protons $Z = 6$ and the number of neutrons $N = 12 - 6 = 6$.The mass defect $\Delta m$ is given by:$\Delta m = [Z \cdot m_H + N \cdot m_n] - M_{nucleus}$$\Delta m = [6 	imes 1.007825 + 6 	imes 1.008665] - 12.00000$$\Delta m = [6.04695 + 6.05199] - 12.00000$$\Delta m = 12.09894 - 12.00000 = 0.09894 	ext{ a.m.u.}$The binding energy $BE = \Delta m 	imes 931.5 	ext{ MeV/a.m.u.}$$BE = 0.09894 	imes 931.5 \approx 92.162 	ext{ MeV}$The binding energy per nucleon is $BE/A = 92.162 / 12 \approx 7.68 	ext{ MeV}$.Comparing with the given options,the closest value is $7.675 	ext{ MeV}$.

The binding energy per nucleon for $^6C^{12}$ nucleus is ......... $MeV$.
(Nuclear mass of $^6C^{12} = 12.00000 \text{ a.m.u.}$
Mass of hydrogen nucleus $= 1.007825 \text{ a.m.u.}$
Mass of neutron $= 1.008665 \text{ a.m.u.}$)

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The binding energy per nucleon for $^6C^{12}$ nucleus is ......... $MeV$.(Nuclear mass of $^6C^{12} = 12.00000 \text{ a.m.u.}$Mass of hydrogen nucleus $= 1.007825 \text{ a.m.u.}$Mass of neutron $= 1.008665 \text{ a.m.u.}$)