Sigma Percentile
LEVELJEE Main

Animated Solution for Physics - Atoms and Nuclei: The energy spectrum of -particles [number as a function of -energy ] emitted from a radioactive source is

Select Answer:

Visualized Solution

  • In -decay, a parent nucleus emits a -particle (electron) and an antineutrino.
  • X \rightarrow Y + e^- + \bar{\nu}

  • The total disintegration energy is shared between the electron and the antineutrino.
  • E_\beta + E_\nu = E_0

  • Since the energy is shared randomly, the kinetic energy of the -particle can be anything from to .
  • 0 \le E_\beta \le E_0

  • The number of particles is zero at and .
  • Due to statistical phase space, the peak occurs at lower energies (around ).

  • The correct graph is continuous, starts at origin, ends at , and is right-skewed.
  • This matches option (c).

The Sigma Insight: Radioactivity

Solution Diagram
Have you ever looked at a graph and realized that it holds the secret to an entirely new, invisible particle? The energy spectrum of -particles is not just a curve on a piece of paper; it is a historical map that led physicists to one of the greatest discoveries of the 20th century. Let's dive into the fascinating world of radioactivity and uncover the story behind this graph.

The Setup

Beta Decay
Imagine a radioactive nucleus sitting quietly. Suddenly, it undergoes -decay. For a long time, scientists believed that this process was simple: a neutron inside the nucleus turns into a proton, and an electron (the -particle) is shot out.
If this were the whole story, the laws of physics would demand something very specific. According to the conservation of energy and momentum, a two-body decay (where a parent nucleus splits into a daughter nucleus and an electron) must result in the electron carrying away a fixed, discrete amount of kinetic energy.
Every single electron emitted from identical nuclei should have the exact same energy, let's call it . If you were to plot the number of particles against their energy , you would expect to see a single, sharp spike at .

The Energy Crisis

But nature had a surprise in store. When experimentalists actually measured the kinetic energy of these emitted -particles, they didn't see a sharp spike. Instead, they saw a broad, continuous curve!
The electrons were coming out with all sorts of energies, ranging from zero all the way up to the maximum energy .
This was a massive crisis in physics. Where was the missing energy going? Was the sacred law of conservation of energy being violated in the quantum realm? Niels Bohr famously suggested that maybe energy conservation was only a statistical law, not an absolute one.

The Neutrino to the Rescue

Enter Wolfgang Pauli. In 1930, he proposed a desperate remedy. He suggested that the missing energy wasn't lost at all. Instead, it was being carried away by a third, invisible particle that was emitted alongside the electron.
This particle had to be electrically neutral (to conserve charge) and incredibly light (to explain the kinematics). Enrico Fermi later named it the "neutrino" (the little neutral one).
Because the decay actually produces three bodies—the daughter nucleus, the electron, and the antineutrino—the total disintegration energy is shared between the electron and the antineutrino.
Since the energy is shared randomly, the electron can have any kinetic energy from zero to , perfectly explaining the continuous spectrum!

The Shape of the Spectrum

Now, let's look at the exact shape of this continuous curve. The number of particles must be zero at (because it's highly unlikely for the electron to get absolutely no energy) and zero at (because it's equally unlikely for the electron to get all the energy while the antineutrino gets none).
But the curve isn't a perfectly symmetric bell shape. Due to the statistical nature of how the energy is shared (the phase space) and the attractive Coulomb force from the positively charged nucleus pulling on the escaping negative electron, the spectrum is skewed.
Most electrons end up with less than half of the total available energy. This means the peak of the curve shifts to the left, occurring at roughly .
When we look at the options provided in the question, we are searching for a graph that is continuous, starts at the origin, ends at , and has a peak skewed to the lower energy side. Option (c) perfectly captures this physical reality.
The next time you see this skewed curve, remember: you are looking at the exact mathematical footprint of the elusive neutrino!

Similar Questions

LEVELJEE Main

Statement I A nucleus having energy decays be emission to daughter nucleus having energy , but rays are emitted with a continuous energy spectrum having end point energy . Statement II To conserve energy and momentum in -decay, atleast three particles must take part in the transformation.

(A)
Statement I is false, Statement II is true
(B)
Statement I is true, Statement II is false
(C)
Statement I is true, Statement II is true; Statement II is the correct explanation of Statement I
(D)
Statement I is true, Statement II is true; Statement II is not the correct explanation of Statement I
LEVELJEE Main

A nucleus with emits the following in a sequence . The of the resulting nucleus is

(A)
76
(B)
78
(C)
82
(D)
74
JEE Advanced 2022
LEVELJEE Main

In a radioactive decay chain reaction, nucleus decays into nucleus. The ratio of the number of to number of particles emitted in this process is_________.

JEE Advanced 2018
LEVELJEE Main

In a radioactive decay chain, nucleus decays to nucleus. Let and be the number of and - particles respectively, emitted in this decay process. Which of the following statements is (are) true?

* Multiple Correct Options
(A)
(B)
(C)
(D)
JEE Main 2021
LEVELJEE Main

If denotes the ratio of the number of nuclei decayed () to the number of nuclei at (), then for a collection of radioactive nuclei, the rate of change of with respect to time is given as [ is the radioactive decay constant]

(A)
(B)
(C)
(D)
LEVELJEE Main

The electron emitted in beta radiation originates from

(A)
inner orbits of atom
(B)
free electrons existing in nuclei
(C)
decay of a neutron in a nucleus
(D)
photon escaping from the nucleus
JEE Main 2010
LEVELJEE Main

A radioactive nucleus (initial mass number and atomic number ) emits -particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

(A)
(B)
(C)
(D)
JEE Advanced 1998
LEVELJEE Advanced

Nuclei of a radioactive element are being produced at a constant rate . The element has a decay constant . At time , there are nuclei of the element. (a) Calculate the number of nuclei of at time . (b) If , calculate the number of nuclei of after one half-life of and also the limiting value of as .

JEE Main 2021
LEVELJEE Main

A sample of a radioactive nucleus disintegrates to another radioactive nucleus , which in turn disintegrates to some other stable nucleus . Plot of a graph showing the variation of number of atoms of nucleus versus time is (Assume that at , there are no atoms in the sample)

(A)
(B)
(C)
(D)
JEE Main 2019
LEVELJEE Main

In a radioactive decay chain, the initial nucleus is . At the end, there are -particles and -particles which are emitted. If the end nucleus is , and are given by

(A)
(B)
(C)
(D)