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Atoms and Nuclei question

2023 · 8 Apr · Shift 1 · Q69
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Atoms and Nuclei question

2023 · 8 Apr · Shift 1 · Q69

JEE MainPhysicsAtoms and NucleiNumerical+4 / −1
A nucleus with mass number 242 and binding energy per nucleon as 7.6 MeV7.6~ \mathrm{MeV}7.6 MeV breaks into two fragment each with mass number 121. If each fragment nucleus has binding energy per nucleon as 8.1 MeV8.1 ~\mathrm{MeV}8.1 MeV, the total gain in binding energy is ‾MeV\underline{\hspace{2cm}}\mathrm{MeV}​MeV.
Numerical answer
View written solutionFree

Correct answer: 121

  1. Given data
  • Initial nucleus mass number: Ai=242A_i = 242Ai​=242
  • Initial binding energy per nucleon: 7.6 MeV7.6\,\text{MeV}7.6MeV
  • It breaks into two fragments, each of mass number 121121121
  • Binding energy per nucleon of each fragment: 8.1 MeV8.1\,\text{MeV}8.1MeV
  1. Total initial binding energy

The total binding energy of the original nucleus is

Bi=242×7.6B_i = 242 \times 7.6Bi​=242×7.6 Bi=1839.2 MeVB_i = 1839.2\,\text{MeV}Bi​=1839.2MeV
  1. Total final binding energy

Each fragment has mass number 121121121, so binding energy of one fragment is

Bf(1)=121×8.1=980.1 MeVB_f^{(1)} = 121 \times 8.1 = 980.1\,\text{MeV}Bf(1)​=121×8.1=980.1MeV

Since there are two identical fragments,

Bf=2×980.1=1960.2 MeVB_f = 2 \times 980.1 = 1960.2\,\text{MeV}Bf​=2×980.1=1960.2MeV
  1. Gain in binding energy

The gain in binding energy is

ΔB=Bf−Bi\Delta B = B_f - B_iΔB=Bf​−Bi​ ΔB=1960.2−1839.2=121.0 MeV\Delta B = 1960.2 - 1839.2 = 121.0\,\text{MeV}ΔB=1960.2−1839.2=121.0MeV
  1. Final answer
121\boxed{121}121​

So, the total gain in binding energy is 121 MeV121\,\text{MeV}121MeV.

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