The $\beta$-decay process, discovered around 1900, is basically the decay of a neutron $(n)$. In the laboratory, a proton $(p)$ and an electron $\left(e^{-}\right)$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. $n \rightarrow p+e^{-}+\bar{v}_{e}$, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino $\left(\bar{v}_{e}\right)$ 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 $0.8 \times 10^{6} \mathrm{eV}$. The kinetic energy carried by the proton is only the recoil energy. What is the maximum energy of the anti-neutrino?

(C) Nearly $0.8 \times 10^{6} \mathrm{eV}$.

(B) Much less than $0.8 \times 10^{6} \mathrm{eV}$.

(D) Much larger than $0.8 \times 10^{6} \mathrm{eV}$.

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Modern Physics

The

β

-decay process, discovered 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 a continuous spectrum. Considering a three-body decay process, i.e.

n \to p + e^{-} + \overset{v}{ˉ}_{e}

, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino

(\overset{v}{ˉ}_{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 1 0^{6} eV

. The kinetic energy carried by the proton is only the recoil energy.

What is the maximum energy of the anti-neutrino?

(A)Zero.
(B)Much less than $0.8 \times 1 0^{6} eV$ .
(C)Nearly $0.8 \times 1 0^{6} eV$ .✓ Correct
(D)Much larger than $0.8 \times 1 0^{6} eV$ .

Answer: (C)

Solution

In the three-body decay

n \to p + e^{-} + \overset{ν}{ˉ}_{e}

, energy and momentum must be conserved. The maximum energy of the anti-neutrino occurs when the electron has minimum kinetic energy (nearly zero). In this case, the anti-neutrino carries away nearly all the available decay energy. Since the maximum electron energy is

0.8 \times 1 0^{6}

eV (when the anti-neutrino has minimum energy), by symmetry and energy conservation, the maximum anti-neutrino energy is also nearly

0.8 \times 1 0^{6}

eV.

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Single correct option

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JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Modern Physics

Solution

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