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 m_{1} and m_{2} only if (m_{1} + m_{2}) < M. Also two light nuclei of masses m_{3} and m_{4} can undergo complete fusion and form a heavy nucleus of mass M' only if (m_{3} + m_{4}) > 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/c^{2})

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

Match List I of the nuclear processes with List II containing parent nucleus and one of the end products of each process and then select the correct answer using the codes given below the lists :

List I | List II | ||
---|---|---|---|

P. | Alpha decay | 1. | $$_8^{15}O \to _7^{15}N + ...$$ |

Q. | $${\beta ^ + }$$ decay | 2. | $$_{91}^{238}U \to _{90}^{234}Th + ...$$ |

R. | Fission | 3. | $$_{83}^{185}Bi \to _{82}^{184}Pb + ...$$ |

S. | Proton emission | 4. | $$_{94}^{239}Pu \to _{57}^{140}La + ...$$ |

The $$\beta$$-decay process, discovered in around 1900, is basically the decay of a neutron (n). In the laboratory, a proton (p) and an electron (e^{$$-$$}) are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has continuous spectrum. Considering a three-body decay process, that is, n $$\to$$ p + e^{$$-$$} + $${\overline v _e}$$, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino ($${\overline v _e}$$) to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is 0.8 $$\times$$ 10^{6} eV. The kinetic energy carried by the proton is only the recoil energy.

What is the maximum energy of the anti-neutrino?

The $$\beta$$-decay process, discovered in around 1900, is basically the decay of a neutron (n). In the laboratory, a proton (p) and an electron (e^{$$-$$}) are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has continuous spectrum. Considering a three-body decay process, that is, n $$\to$$ p + e^{$$-$$} + $${\overline v _e}$$, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino ($${\overline v _e}$$) to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is 0.8 $$\times$$ 10^{6} eV. The kinetic energy carried by the proton is only the recoil energy.

If the anti-neutrino had a mass of 3 eV/c^{2} (where c is the speed of light) instead of zero mass, what should be the range of the kinetic energy, K, of the electron?