Question

The binding energy per nucleon for a 6c12 nucleus is

Answer 1

Answer:

Nuclear binding energy is the energy required to split a nucleus of an atom into its component parts: protons and neutrons, or, collectively, the nucleons.

Answer 2

The binding energy per nucleon is 6.65 MeV

Explanation:

Nucleons are the sub-atomic particles that are present in the nucleus of an atom. The sub-atomic particles present in the nucleus are protons and neutrons.

We are given a nucleus having representation:  $_{6}^{12}\textrm{C}$

To calculate the mass defect of the nucleus, we use the equation:

$\Delta m=[(n_p\times m_p)+(n_n\times m_n)-M$

where,

$n_p$ = number of protons = 6  

$m_p$ = mass of one proton = 1.00728 amu

$n_n$ = number of neutrons = 12 - 6 = 6

$m_n$ = mass of one neutron = 1.00866 amu

M = nuclear mass = 12.01 amu  

Putting values in above equation, we get:

$\Delta m=[(6\times 1.00728)+(6\times 1.00866)]-12.01\\\\\Delta m=0.08564amu$

To calculate the binding energy of the nucleus, we use the equation:

$E=\Delta mc^2\\\\E=(0.08564)\times c^2$

$E=(0.08564u)\times (931.5MeV)$    (Conversion factor:  $1u=931.5MeV/c^2$  )

$E=79.774MeV$

Number of nucleons in $_{6}^{12}\textrm{C}$ atom = 12

To calculate the binding energy per nucleon, we divide the binding energy by the number of nucleons, we get:

$\text{Binding energy per nucleon}=\frac{\text{Binding energy}}{\text{Nucleons}}$

$\text{Binding energy per nucleon}=\frac{79.774}{12}=6.65MeV$

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