Question

Consider the nuclear fission

Ne²⁰ → 2He⁴ + C¹²

Given that the binding energy/nucleon of Ne²⁰, He⁴ and C¹² are respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement.
1. Energy of 3.6 MeV will be released
2. Energy of 12.4 MeV will be supplied
3. 8.3 MeV energy will be released
4. Energy of 9.72 MeV has to be supplied

Answer 1

Correct Answer - Option 4 : Energy of 9.72 MeV has to be supplied

Note:

For this question in Jee paper, there was no correct option given, hence one of the options is replaced by the correct one.

 

Concept:

Energy absorbed or released in a nuclear reaction is given by;

ΔQ = Binding energy of products – Binding energy of reactants.

If energy is absorbed, ΔQ is negative and if it is positive then energy is released.

Also, Binding energy = Binding energy per nucleon × number of nucleons.

Calculation:

Given reaction is

Ne²⁰ → 2He⁴ + C¹²

Binding energy per nucleon of Ne²⁰ is 8.03 MeV

Number of nucleon in Ne²⁰ is 20

Binding energy per nucleon of He⁴ is 7.07 MeV

Number of nucleon in He⁴ is 4

Binding energy per nucleon of C¹² is 7.86 MeV

Number of nucleon in C¹² is 12

Here, binding energy of products i.e., 2(He)⁴ + C¹² is

= 2 × (B.E of He⁴) + (B.E of C¹²)

= 2(4 × 7.07) + (12 × 7.86)

= 2 × 28.28 + 94.32

= 56.56 + 94.32

= 150.88 MeV

And binding energy of reactants, i.e., Ne²⁰ is

= 20 × 8.03 = 160.6 MeV

So, ΔQ = (B.E)_Products – (B.E)_reactants

= 150.88 – 160.6 = -9.72 MeV

As ΔQ is negative,

∴ Energy of 9.72 MeV is absorbed in the reaction.