Question

Nucleus $A$ having $Z=17$ and equal number of protons and neutrons has $12 MeV$ binding energy per nucleonAnother nucleus $B$ of $Z=12$ has total 26 nucleons and $18 MeV$ binding energy per nucleons The difference of binding energy of $B$ and $A$ will be ___$MeV$

Hint:

Remember that the mass number is the sum of protons and neutrons. The total binding energy is the product of the binding energy per nucleon and the mass number.

Answer 1

Correct Answer: 6

Step 1: Calculate the Mass Number of Nucleus A

Nucleus A has \(Z = 17\) (number of protons). Since it has an equal number of protons and neutrons, the number of neutrons is also 17. The mass number (\(A\)) is the sum of protons and neutrons:

\[ A = 17 + 17 = 34 \]

Step 2: Calculate the Total Binding Energy of Nucleus A

The binding energy per nucleon for nucleus A is \(1.2 \, \text{MeV}\). The total binding energy is the product of the binding energy per nucleon and the mass number:

\[ BE_A = 1.2 \times 34 = 40.8 \, \text{MeV} \]

Step 3: Calculate the Total Binding Energy of Nucleus B

Nucleus B has 26 nucleons, and the binding energy per nucleon is \(1.8 \, \text{MeV}\):

\[ BE_B = 1.8 \times 26 = 46.8 \, \text{MeV} \]

Step 4: Calculate the Difference in Binding Energies

The difference in binding energies is:

\[ \Delta BE = BE_B - BE_A = 46.8 - 40.8 = 6 \, \text{MeV} \]

Conclusion: The difference in binding energy is \(6 \, \text{MeV}\).

Answer 2

The correct answer is 6.
For A mass number $=34$
Total binding energy $=1.2\times 34=40.8MeV$
For $B$ mass number $=26$
total binding energy $=1.8\times 26MeV$
$=46.8MeV$
Difference of $BE=6MeV$