Skip to content
dronacharya.ai
← IIT-JEE 2012 · Paper 2
PhysicsSingle correct+3 / -1Question 12 of 60

IIT-JEE 2012 · Paper 2, Physics question 12

Passage

Paragraph for Questions 11 and 12

The β\beta-decay process, discovered around 1900, is basically the decay of a neutron (n)(n). In the laboratory, a proton (p)(p) and an electron (e−)(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 a continuous spectrum. Considering a three-body decay process, i.e. n→p+e−+νˉen\to p+e^-+\bar\nu_e, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino (νˉe)(\bar\nu_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 0.8×1060.8\times10^6 eV. The kinetic energy carried by the proton is only the recoil energy.

If the anti-neutrino had a mass of 3 eV/c2c^2 (where cc is the speed of light) instead of zero mass, what should be the range of the kinetic energy, KK, of the electron?

Choose one option
Answer and worked solution

Official answer: D

With a 3 eV/c² antineutrino, 3 eV of the released energy must go into its rest mass, so the electron can never take the full 0.8×1060.8\times10^6 eV: K<0.8×106K<0.8\times10^6 eV.

Lower limit. The electron can still be produced (almost) at rest, with the energy carried by the antineutrino and the proton: K≥0K\ge0.

(D) 0≤K<0.8×1060\le K<0.8\times10^6 eV

Topics:Radioactive decay
← Previous questionNext question →