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Determine the binding energy per nucleon of an mg-24 nucleus. the mg-24 nucleus has a mass of 24.30506. a proton has a mass of 1.00728 amu, a neutron has a mass of 1.008665 amu, and 1 amu is equivalent to 931 mev of energy.

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SonOfZeus

The mass of Mg-24 is 24.30506 amu, it contains 12 protons and 12 neutrons.

Theoretical mass of Mg-24:

The theoretical mass of Mg-24 is:

Hydrogen atom mass = 12 × 1.00728 amu = 12.0874 amu

Neutron mass = 12 x 1.008665 amu = 12.104 amu

Theoretical mass = Hydrogen atom mass + Neutron mass = 24.1913 amu

Note that the mass defect is:

Mass defect = Actual mass - Theoretical mass : 24.30506 amu- 24.1913 amu= 0.11376 amu

Calculating the binding energy per nucleon:

[tex]\frac{B.E.}{nucleon}=\frac{(0.11376amu)(931Mev/amu}{24nucleons}  = 4.41294 Mev/nucleon[/tex]

So approximately 4.41294 Mev/necleon


hamzaahmeds

This question involves the concepts of binding energy, mass defect, and nucleons.

The Binding energy per nucleon of an Mg-24 nucleus is "4.4 MeV/nucleon".

The number of nucloens is the total number of protons and neutrons in a nucleus of an atom. Therefore,

n = No. of nucleons = 12 protons + 12 nuetrons = 24

Now, the mass defect can be calculated as follows:

[tex]\Delta m = M- m_pn_p+m_nn_n[/tex]

where,

Δm = mass defect = ?

[tex]m_p[/tex] = mass of proton = 1.00728 amu

[tex]n_p[/tex] = no. of protons = 12

[tex]m_n[/tex] = mass of neutron = 1.008665 amu

[tex]n_n[/tex] = no. of netron = 12

M = mass of Mg-24 = 24.30506 amu

Therefore,

[tex]\Delta m =24.30506\ amu -(12)(1.00728\ amu)+(12)(1.008665\ amu)[/tex]

Δm = 0.11372 amu[tex](\frac{931\ MeV}{1\ amu})[/tex] = 105.87 MeV

Now, the binding energy per nucleon will be:

[tex]\frac{Binding\ Energy}{nucleon}=\frac{105.87\ MeV}{24\ nucleon}\\\\\frac{Binding\ Energy}{nucleon}=4.4\ MeV/nucleon[/tex]

Learn more about binding energy here:

https://brainly.com/question/10095561?referrer=searchResults

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