Question 73: Atoms & Nuclei

Question

The mass of a nucleus $$_Z^AX$$ is less than the sum of the masses of (A-Z) number of neutrons and Z 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 M can break into two light nuclei of masses m1 and m2 only if (m1 + m2)  M'. The masses of some neutral atoms are given in the table below :

$$_1^1H$$
1.007825 u
$$_1^2H$$
2.014102 u

$$_3^6Li$$
6.015123 u
$$_3^7Li$$
7.016004 u

$$_{64}^{152}Gd$$
151.919803 u
$$_{82}^{206}Pb$$
205.974455 u

$$_1^3H$$
3.016050 u
$$_2^4He$$
4.002603 u

$$_{30}^{70}Zn$$
69.925325 u
$$_{34}^{82}Se$$
81.916709 u

$$_{83}^{209}Bi$$
208.980388 u
$$_{84}^{210}Po$$
209.982876 u

(1 u = 932 MeV/c2)

Accepted Answer

We have
$$m(_1^2H) + m(_2^4He) = 2.014102 + 4.002603 = 6.016705\,u$$
$$m(_3^6Li) = 6.015123\,u$$
$${m_1} + {m_2} > M$$
Thus, option (A) is wrong.
$$m(_1^1H) + m(_{83}^{209}Bi) = 1.007825\,u + 208.980388\,u = 209.988213\,u$$
$$m(_{84}^{210}Po) = 209.982876\,u$$
$${m_1} + {m_2} > M$$
Thus, option (B) is wrong.
$$m(_1^2H) + m(_2^4He) = 2.014102\,u + 4.002603\,u = 6.016705\,u$$
$$_3^6Li = 6.015123\,u$$
$$({m_3} + {m_4}) > M'$$
Thus, option (C) is correct. Therefore, deuteron and alpha particle can go complete fusion.
$$m(_{30}^{70}Zn) + _{34}^{82}Se = 69.925325\,u + 81.916709\,u = 151.842034\,u$$
$$_{64}^{152}Gd = 151.919803\,u$$
$${m_3} + {m_4} < M'$$
Thus, option (D) is wrong.

Answer

the nucleus $$_3^6Li$$ can emit an alpha particle.

Answer

the nucleus $$_{84}^{210}Po$$ can emit a proton.

Answer

deuteron and alpha particle can undergo complete fusion.

Answer

the nuclei $$_{30}^{70}Zn$$ and $$_{34}^{82}Se$$ can undergo complete fusion.

Atoms & Nuclei

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