Question

The nuclear mass of 37cl is 36.9566 amu. calculate the binding energy per nucleon for 37cl.

Answer 1

Answer: The binding energy per nucleon is
`1.4* 10^(-12)J`

Step-by-step explanation:

Nucleons are defined as the sub-atomic particles which are present in the nucleus of an atom. Nucleons are protons and neutrons.

We are given a nucleus having representation:
`_(17)^(37)\textrm{Cl}`

Number of protons = 17

Number of neutrons = 37 - 17 = 20

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

`\Delta m=[(n_p* m_p)+(n_n* m_n)-M`

where,

`n_p` = number of protons = 17

`m_p` = mass of one proton = 1.00728 amu

`n_n` = number of neutrons = 20

`m_n` = mass of one neutron = 1.00866 amu

M = nuclear mass = 36.9566 amu

Putting values in above equation, we get:

`\Delta m=[(17* 1.00728)+(20* 1.00866)]-36.9566\\\\\Delta m=0.34036amu`

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

`E=\Delta mc^2\\E=(0.34036u)* c^2`

`E=(0.34036u)* (931.5MeV)` (Conversion factor:
`1u=931.5MeV/c^2` )

`E=317.04MeV=507.264* 10^(-13)J` (Conversion factor:
`1MeV=1.6* 10^(-13)J` )

Number of nucleons in
`_(17)^(37)\textrm{Cl}` atom = 37

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}=(507.26* 10^(-13)J)/(37)=1.4* 10^(-12)J`

Hence, the binding energy per nucleon is
`1.4* 10^(-12)J`

Answer 2

The 17 Cl 37 nucleus has 37 nucleons. Thus, making the energy per nucleon is 1.6944 x 10^-19 J/37 nucleons. Energy = 5.09x10^-11 J. So, (5.09x10^-11)/37 nucleons you get a 1.38x10^-12 J/nucleon as an answer.

I hope this helps you with your problem.