Sigma Percentile
JEE Advanced 2012
LEVELJEE Advanced

Animated Solution for Physics - Atoms and Nuclei: Comprehension Passage

The -decay process, discovered around 1900, is basically the decay of a neutron (). In the laboratory, a proton () and an electron () 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., , around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino () 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 eV. The kinetic energy carried by the proton is only the recoil energy.
Question 1:

If the anti-neutrino had a mass of (where is the speed of light) instead of zero mass, what should be the range of the kinetic energy , of the electron?

Select Answer:

Question 2:

What is the maximum energy of the anti-neutrino?

Select Answer:

Visualized Solution

The Sigma Insight: Radioactivity

Solution Diagram

The Mystery of Beta Decay

Imagine you are a physicist in the 1920s. You are observing beta decay, where a neutron decays into a proton and an electron. According to the laws of physics, if a particle at rest decays into two particles, they must fly apart with specific, constant energies to conserve both momentum and energy. But experiments showed something baffling: the emitted electrons had a continuous range of energies! It seemed as if energy was vanishing into thin air.
To save the sacred law of energy conservation, Wolfgang Pauli proposed a radical idea in 1930: a "ghost" particle was carrying away the missing energy. This particle, later named the neutrino (or anti-neutrino in this specific case), was neutral, incredibly light, and interacted so weakly with matter that it was invisible to detectors of that era.

Energy Conservation in Action

Let's break down the physics. When a neutron decays, the total energy released is called the -value. In our problem, this is given as . This energy must be shared among the products: the proton, the electron, and the anti-neutrino.
Because the proton is thousands of times more massive than the electron and the anti-neutrino, it barely moves. Its recoil kinetic energy () is practically zero. Thus, the -value is essentially shared between the electron and the anti-neutrino:

The Massive Anti-neutrino

Now, let's tackle the first question. What if the anti-neutrino isn't perfectly massless? Suppose it has a tiny mass of . According to Einstein's famous equation, , any particle with mass has an intrinsic rest mass energy. For our anti-neutrino, this is .
This means that even if the anti-neutrino is completely stationary, it must consume of the available -value just to exist!
Because the anti-neutrino is stealing at least , the electron can never have the full . Its maximum possible kinetic energy is strictly less than the -value:
Since kinetic energy cannot be negative, the electron's kinetic energy must fall in the range:

The Maximum Energy of the Anti-neutrino

Moving to the second question, we want to find the absolute maximum energy the anti-neutrino can possess. Since the total energy is fixed, the anti-neutrino gets the most energy when the electron gets the least.
The minimum kinetic energy the electron can have is zero (it just pops into existence and sits there). If , then the anti-neutrino runs away with almost all the available energy:
This beautiful interplay of energy and momentum conservation not only solves our problem but also highlights how physicists deduce the properties of the invisible quantum world!

Similar Questions

JEE Advanced 2019
LEVELJEE Advanced

Suppose a nucleus at rest and in ground state undergoes -decay to a nucleus in its excited state. The kinetic energy of the emitted particle is found to be . nucleus then goes to its ground state by -decay. The energy of the emitted -photon is _______ , [Given: atomic mass of , atomic mass of , atomic mass of particle = , , is speed of the light]

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Statement I is false, Statement II is true
(B)
Statement I is true, Statement II is false
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(B)
(C)
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A nucleus with mass number 220 initially at rest emits an -particle. If the value of the reaction is 5.5 MeV, calculate the kinetic energy of the -particle

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During a negative beta decay,

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(B)
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JEE Advanced 1986
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There is a stream of neutrons with a kinetic energy of . If the half-life of neutrons is , what fraction of neutrons will decay before they travel a distance of ?

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The electron emitted in beta radiation originates from

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A nucleus , initially at rest, undergoes alpha-decay according to the equation. (a) Find the values of and in the above process. (b) The alpha particle produced in the above process is found to move in a circular track of radius in a uniform magnetic field of . Find the energy (in MeV) released during the process and the binding energy of the parent nucleus . Given that ; ; ; .

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In the nuclear process, , stands for ........ .