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← JEE Advanced 2013 · Paper 2
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JEE Advanced 2013 · Paper 2, Physics question 12

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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 kinetic energy (in keV) of the alpha particle, when the nucleus 84210Po^{210}_{84}\mathrm{Po} at rest undergoes alpha decay, is

Choose one option
Answer and worked solution

Official answer: A

Q-value. 210^{210}Po → 206^{206}Pb + 4^4He:
Δm=209.982876−205.974455−4.002603=0.005818 u,Q=0.005818×932=5.422 MeV.\Delta m=209.982876-205.974455-4.002603=0.005818\ \text{u},\quad Q=0.005818\times932=5.422\ \text{MeV}.

Alpha share (momentum conservation): Kα=Q206210=5.319K_\alpha=Q\dfrac{206}{210}=5.319 MeV = 5319 keV.

(A)