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JEE Advanced 2013 · Paper 2, Physics question 11

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Paragraph for Questions 11 and 12

The mass of a nucleus ZAX^A_Z\mathrm X is less than the sum of the masses of (A−Z)(A-Z) number of neutrons and ZZ number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass MM can break into two light nuclei of masses m1m_1 and m2m_2 only if (m1+m2)<M(m_1+m_2)<M. Also two light nuclei of masses m3m_3 and m4m_4 can undergo complete fusion and form a heavy nucleus of mass M′M' only if (m3+m4)>M′(m_3+m_4)>M'. The masses of some neutral atoms are given in the table below:

Atom Mass Atom Mass
11H^1_1\mathrm H 1.007825 u 12H^2_1\mathrm H 2.014102 u
13H^3_1\mathrm H 3.016050 u 24He^4_2\mathrm{He} 4.002603 u
36Li^6_3\mathrm{Li} 6.015123 u 37Li^7_3\mathrm{Li} 7.016004 u
3070Zn^{70}_{30}\mathrm{Zn} 69.925325 u 3482Se^{82}_{34}\mathrm{Se} 81.916709 u
64152Gd^{152}_{64}\mathrm{Gd} 151.919803 u 82206Pb^{206}_{82}\mathrm{Pb} 205.974455 u
83209Bi^{209}_{83}\mathrm{Bi} 208.980388 u 84210Po^{210}_{84}\mathrm{Po} 209.982876 u

(1 u = 932 MeV/c²)

The correct statement is

Choose one option
Answer and worked solution

Official answer: C

Check each mass balance (u):

  • (A) 6^6Li → 4^4He + 2^2H: 4.002603+2.014102=6.016705>6.0151234.002603+2.014102=6.016705>6.015123. Not allowed.
  • (B) 210^{210}Po → 209^{209}Bi + 1^1H: 208.980388+1.007825=209.988213>209.982876208.980388+1.007825=209.988213>209.982876. Not allowed.
  • (C) 2^2H + 4^4He → 6^6Li: 6.016705>6.0151236.016705>6.015123. Fusion allowed. ✓
  • (D) 70^{70}Zn + 82^{82}Se → 152^{152}Gd: 151.842034<151.919803151.842034<151.919803. Not allowed.

(C)

Topics:Mass defect and binding energyNuclear fission and fusion
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