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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
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
Step 1: Calculate the Mass Defect
Answer
To calculate the binding energy, we first need to find the mass defect ( \Delta m ). The mass defect is given by the formula: [ \Delta m = Z \cdot m_p + N \cdot m_n - M ] where:\
- ( Z ) is the atomic number (number of protons)\
- ( N ) is the number of neutrons\
- ( m_p ) is the mass of a proton (approximately ( 1.00728 , u ))\
- ( m_n ) is the mass of a neutron (approximately ( 1.00866 , u ))\
- ( M ) is the actual mass of the nucleus (in this case, ( 58.93320 , u ))
For ( {}^{59}_{27}Co ): - ( Z = 27 ) (protons)\
- ( N = 59 - 27 = 32 ) (neutrons)
This gives us:
[\Delta m = (27 \cdot 1.00728 + 32 \cdot 1.00866 - 58.93320) , u]
Step 2
Step 2: Calculate the Binding Energy
Answer
Next, we convert the mass defect to energy using the mass-energy equivalence formula: [ E = \Delta m \cdot c^2 ] In atomic mass units, 1 u corresponds to about 931.5 MeV. Hence, the binding energy is calculated as follows: [E = \Delta m \cdot 931.5 , \text{MeV/u}] This will yield the binding energy of the nucleus in MeV.