Nuclei

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Common Misconceptions

3 min read

- Let us bust some common misconceptions

What determines the harmfulness of radiation: Ionizing Power or Penetrating power?

Ionizing Power:When a radiation particle interacts with atoms, the interaction can cause the atom to lose electrons and thus become ionized. Because of the large mass of the alpha particle which is nothing but a Helium nucleus, it has the highest ionizing power and the greatest ability to damage tissue.

Penetrating Power:The capacity of a radiation beam to penetrate through a medium is its penetrating power. Because of the large size of alpha particles it can be stopped by a sheet of paper. Hence, bigger the size of radiation particles lesser is the penetrating power.

The following table gives a quantitative comparison of alpha beta and gamma particles in terms of their ionizing power and penetrating power.

We have understood that alpha particles have thehighest ionizing power and gammarays have thehighest penetrating power.So, which one do you think is more harmful?

Gamma rays are the most harmful source of External Radiation, as a gamma source lying close to you can cause radiation damage depending on the extent of exposure.Alpha rays are on the other hand are the most dangerous source of Internal Radiationif an alpha source manages to enter our body. Alpha particles can be easily blocked on the outside but once inside the body, it can be extremely dangerous.

In alpha decay, why does the radioactive nucleus emit an alpha particle and not four separate nucleons?

In the process of spontaneous alpha decay of Uranium it decays into a Thorium nucleus and an alpha particle.Why cant it decay by emitting 4 separate nucleons?

For the reaction

_{92}^{238} U \to_{90}^{234} T h +_{2}^{4} H e

the Q value is found to be positive.

The energy of the reaction or the Q-value is :

Q = [m (_{92}^{238} U) - m (_{90}^{234} T h) - m (_{2}^{4} H e)] c^{2} > 0

For spontaneous decay, the energy of reaction must be positive as the nucleus must release energy.In this case the released energy appears as kinetic energy in alpha particle.

However, for the proposed reaction where 4 nucleons are released the Q value turns out to be negative which isnot possible as it violates the law of conservation of energy.

In any decay process, the conservation of energy, conservation of linear momentum and conservation of angular momentum must be obeyed.

Why do heavy nuclei have disproportionately higher number of neutrons?

You already know a nucleus consists of protons and neutrons. The mass number of a nucleus is the number of nucleons or the number of protons(Z) and number of neutrons(N). The number of protons and neutrons for most small nuclei is somewhat close together if not equal. However in naturally occurring heavy elements the number of neutrons is comparatively higher as compared to protons, as you can see in the graph below.

Why do you think this happens?

The nucleons inside a nucleus are held together by strong attractive nuclear forces and these forces must counter the repulsive electrostatic forcebetween the protons within a nucleus. Let us draw a comparison between the electrostatic and nuclear force experienced by he nucleons.

From the given plot, we can see, nuclear force is a relatively short range force and is repulsive at distances less that 0.8fm and attractive at distances greater than it, but it drops to negligible values at a range of 3fm. However electrostatic forces act between protons at even higher distances. A stable nucleus can form when the attractive forces between nucleons compensate for the repulsive, long-range electrostatic forces between all protons in the nucleus otherwise the repulsive force between the protons will break the nucleus apart.