Question

Nucleus A having Z = 17 and equal number of protons and neutrons has 1.2MeV binding energy per nucleon. Another nucleus B of Z = 12 has total 26 nucleons and 1.8 MeV binding energy per nucleons. The difference of binding energy of B and A will be

• A. 6MeV
• B. 12MeV
• C. 18MeV
• D. 24MeV

Answer 1

Answer: (A)

6MeV

Answer 2

Here's how to calculate the difference in binding energy:

1. Calculate the total binding energy of nucleus A:

   • Mass number of A ($A_A$) = Number of protons + Number of neutrons = 17 + 17 = 34
   • Total binding energy of A ($BE_A$) = $A_A$ * (Binding energy per nucleon) = 34 * 1.2 MeV = 40.8 MeV

2. Calculate the total binding energy of nucleus B:

   • Mass number of B ($A_B$) = 26
   • Total binding energy of B ($BE_B$) = $A_B$ * (Binding energy per nucleon) = 26 * 1.8 MeV = 46.8 MeV

3. Calculate the difference in binding energy:

   • Difference = $BE_B$ - $BE_A$ = 46.8 MeV - 40.8 MeV = 6.0 MeV

Therefore, the difference in binding energy between nucleus B and nucleus A is 6 MeV.