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Calculate the binding energy, in MeV, of a nucleus of $^{59}_{27}$Co - AQA - A-Level Physics - Question 5 - 2017 - Paper 2

Question 5

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Calculate the binding energy, in MeV, of a nucleus of $^{59}_{27}$Co. nuclear mass of $^{59}_{27}$Co = 58.93320 u

Worked Solution & Example Answer:Calculate the binding energy, in MeV, of a nucleus of $^{59}_{27}$Co - AQA - A-Level Physics - Question 5 - 2017 - Paper 2

Step 1

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Answer

To calculate the binding energy, we will use the mass defect formula:

a) Calculate the total mass of the individual nucleons:

Each proton has a mass of approximately 1.007276 u.
Each neutron has a mass of approximately 1.008665 u.
For $^{59}_{27}$ Co, there are 27 protons and 32 neutrons (59 - 27 = 32). Therefore:

$M_{tot} = (27 imes 1.007276 ext{ u}) + (32 imes 1.008665 ext{ u})$ $M_{tot} = 27.197432 + 32.27728 = 59.474712 ext{ u}$

b) The mass defect ( $ext{Δm}$ ) is found by subtracting the actual nuclear mass from the total mass of the nucleons:

$ext{Δm} = 59.474712 ext{ u} - 58.93320 ext{ u} = 0.541512 ext{ u}$

c) Convert the mass defect to energy using Einstein's relation $E = ext{Δm} imes c^2$ , where $c^2$ is approximately 931.5 MeV/u:

$E_{binding} = 0.541512 imes 931.5 ext{ MeV/u} ext{ (as } c^2 = 931.5 ext{ MeV/u)}$ $E_{binding} ext{ = } 504.5 ext{ MeV}$

Thus, the binding energy of the nucleus is approximately 504.5 MeV.