alpha decay

Alpha decay

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Alpha decay

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Glow-in-the-dark toys and watch dials work on the principle that when exposed to light, they absorb it and when brought into a dark room, re-emit the light gradually over a few minutes or even hours. But they glow dimmer and dimmer with time and ultimately the light fades away completely. This makes sense because we can think in terms of the object glowing so long as it still contains the light it had absorbed. Once all the stored light has been emitted out, there’s no more glow.

But imagine that you came across an object that’d never stop glowing or doesn’t even dim with time. It has no batteries or wires connected to it so as to supply it energy. Nor is it being heated in any way. It just glows, constantly, for years. What would you make of it? A similar situation was encountered by a French physicist, Henri Becquerel towards the end of the 19th C. Only it was a little less dramatic, in that he observed uranium salts forming images on photographic plates (rather than see the salts glow visibly). This meant that the salts were emitting some form of radiation. But the source of this stored energy was unknown. The salts continued radiating even if they were kept locked in a dark room for months. Becquerel had stumblled across a bizzare happening, one that even seemed to defy the sacred law of energy conservation (that energy cannot be created out of nowhere).

This strange phenomenon caught the curiosity of the Curies too. While still puzzled by the origin of these radiations, Marie and Pierre Curie together discovered that another element Thorium too behaved similarly. The radiations became known as Becquerel rays and the Curies named this phenomenon \textbf{radioactivity}. Fascinated by the mystery, they went on further to even discover two new elements – Polonium and Radium – that emitted the Becquerel rays.

It was not until Lord Rutherford and his coworkers perfomed rigorous experiments that the precise nature of these mysterious radiations were known. It was found that the Becquerel rays are actually a mixture of three different radiations: the alpha ($\alpha$), the beta ($\beta$) and the gamma ($\gamma$), named in increasing order of their penetrating powers. The alpha rays consist of a stream of helium nuclie (each comprising of two protons and two neutrons), the beta rays are simply streams of high energy electrons and the gamma rays are energetic photons that have the greatest penetrating power.

But none of this explains why the radiation is emitted in the first place and where it is coming from. The idea is that these radiations are nature’s way of restoring stability. Atomic nuclie are essentially nucleons (protons and neutrons) stuck together through the strong nuclear force. The protons are held together despite their electrical repulsion (positive-positive repel). The nuclear force is strong enough to overcome this repulsion but it falls weak very quickly with distance. If we have a nucleus that is quite large, the distance between two protons that are situated at diametrically opposite ends of the nucleus could be large enough so that the nuclear force would hardly be attracting them together while the electrical repulsion between the protons that stays strong at these distances, threatens to tear the nucleus apart.

This is a very unstable situation for the nucleus. One of the ways to bring back stability is to get rid of a chunck of the nucleus so that the nuclear size reduces and all the protons that are left are within good reach of the nuclear force. This chunck of the nucleus that appears to break free and fly out is the alpha particle (a single bunch of two protons and two neutrons). The overall effect is to reduce the number of nucleons by four.

Say ${}_Z^AX$ is the unstable nucleus (called the parent nucleus), where $Z$ (atomic number) is the number of protons and $A$ (mass number) is the total number of nucleons (protons + neutrons). We can represent this as the reaction:

\[{}_Z^A\mathrm{X}\rightarrow {}_{Z-2}^{A-4}\mathrm{Y}+{}^4_2\mathrm{He}+\mathrm{Q}\]

The result of the alpha decay, besides the ejection of the alpha particle, is that the parent nucleus transforms into a different nucleus (called the daughter), for it is the atomic number that distinguishes one nucleus from another. This is accompanied with the release of some energy $Q$, called the disintegration energy.

To relate this with our earlier mentioning of the very reason for the decay to happen at all, we must observe that $Y$, the daughter nucleus, is more stable that $X$, the parent. Lesser nucleons means smaller nucleus which in turn ensures a better reach of the short-range nuclear force, reducing the threat from the electrical repulsions.

It must however be noted here that alpha decay is not as straightforward as it is to say that a chunck of a heavy nucleus breaks free. In fact if observed closely, you’ll find that the alpha particle does not posses enough energy to eject itself from the parent nucleus. But down at those scales, the alpha particle behaves like a wave according to quantum mechanics and hence is capable of doing what classical particles can’t.

The alpha particle while inside the nucleus is free to move about within it and hence knocks several times at the nuclear boundary (somewhat like gas molecules bumping into the walls of the container). Quantum mechanics allows for some probability of the alpha particle crossing over the boundary – called the tunneling effect. And hence we observe alpha particles shooting out of unstable nuclei. No wonder how puzzled Becquerel and the Curies were, witnessing this phenomenon in times when quantum mechanics was completely non-existent.

Now with the emission of the alpha particle, there is a release of some energy, called the disintegration energy $Q$. This is a result of the mass difference between the products and the reactants of the decay. It so happens that the rest mass (measured when you see the mass at rest) of the parent nucleus is larger than the rest mass of the daughter nucleus plus that of the alpha particle. This missing mass manifests as the energy $Q$, according to the Einstein’s mass-energy equivalence relation, $Q=\Delta mc^2$. In fact, the whole purpose of the alpha decay can be attributed to the release of this energy, hence reducing the nuclear energy, making it more stable.

To put it more clearly, if $m_X$ is the rest mass of the parent nucleus, $m_Y$, that of the daugther nucleus and $m_\alpha$, of the alpha particle, it turns out that,

\[m_X\neq m_Y+m_\alpha\]

Rather, \[m_X>m_Y+m_\alpha\]

The missing mass $\Delta m=m_X-(m_Y+m_\alpha)$ shows up as the energy $Q$ according to \[Q=(m_X-m_Y-m_\alpha)c^2\]

Most of this energy appears in the form of the kinetic energy of the ejected alpha particle. The rest accounts for the tiny energy with which the daughter nucleus recoils (to conserve momentum). This is much like the backward recoil of the rifle when the bullet is fired forwards.

To calculate the kinetic energy of the ejected alpha particles, let us first use the energy conservation law that tells us that the disintegration energy $Q$ must be a sum of the kinetic energies of the alpha particle and the daughter nucleus. I.e.,

\[Q=K_Y+K_{\alpha}\]

Using the classical expression for kinetic energy in terms of momentum, $K=p^2/2m$, we can write,

\[Q={p_Y^2\over 2M_Y}+{p_{\alpha}^2\over 2M_{\alpha}}\]

But then the momentum conservation law requires that the alpha particle and the daughter nucleus have equal and opposite momenta (assuming the parent nucleus was at rest initially). I.e.,

\[p_Y=p_{\alpha}\]

Making this substitution in th expression for energy,

\begin{align*}

Q&={p_{\alpha}^2\over 2M_Y}+{p_{\alpha}^2\over 2M_{\alpha}}\\

&={p_{\alpha}^2\over 2}\left({1\over M_Y}+{1\over M_{\alpha}}\right)\\

&={p_{\alpha}^2\over 2}\left({M_{\alpha}+M_Y\over M_{\alpha}M_Y}\right)\\

&={p_{\alpha}^2\over 2M_{\alpha}}\left({M_{\alpha}+M_Y\over M_Y}\right)\\

&=K_{\alpha}\left({M_{\alpha}+M_Y\over M_Y}\right)

\end{align*}

Or, rearranging,

\begin{equation}

K_{\alpha}=\left({M_Y\over M_{\alpha}+M_Y}\right)Q \label{eq: Kinetic energy}

\end{equation}

Hence knowing the masses of the daughter nucleus $M_Y$, the alpha particle, $M_{\alpha}$ and the disintegration energy, $Q$, we can compute the kinetic energy of the alpha particle. We could also rewrite this expression in terms of the mass number $A$ of the parent nucleus. Since the daughter nucleus is short of four nucleons that have discharged in the form of the alpha particle, its mass $M_Y\approx(A-4)m$, where $m=m_p\approx m_n$ is the mass of one nucleon (the mass of the neutron is very slightly greater than that of the proton, hence the approximation). Similarly, $M_{\alpha}+M_Y\approx 4m+(A-4)m=Am$. Hence,

\begin{equation*}

K_{\alpha}\approx \left({A-4\over A}\right)Q

\end{equation*}

Now with alpha particles ejecting off the parent nucleus, despite the reduction in the mass and hence size of the nucleus, the daughter nucleus needn’t be completely stable. In fact, in most cases the daughter nucleus undergoes further radioactive decays until the most stable configuration is reached.

As a concrete example, consider the Uranium-238 nucleus. It comprises of a gigantic 92 protons and 146 neutrons. When it undergoes an alpha decay, it transmutes to a Thorium-234 nucleus.

\[{}^{238}_{92}\mathrm{U}\rightarrow {}^{234}_{90}\mathrm{Th}+{}^4_2\mathrm{He}+\mathrm{Q}\]

The Thorium nucleus is in itself radioactive and further (beta) decays into Protactinium, and so on. The decays continue happening in succession until the nucleus of lead, comprising of 206 nucleons is reached. Lead is not radioactive and never decays further.

To calculate the value of Q here, we use the expression,

\[Q=(M_X-M_Y-M_\alpha)c^2\]

We should here note that the mass of the parent nucleus is not just the sum of the masses of its constituent nucleons, $M_X\neq 92m_p+146m_n$. The nuclear binding energy and the electric repulsion between the protons affects the overall mass of the nucleus. Without digressing into those calculations, we for now, assume to know the values: $M_X=238.050770$ u, $M_Y=234.043583$ u and $M_{\alpha}=4.002603$ u, where $1u=1.6605\times 10^{-27}$ is called the atomic mass unit and is the average of the masses of the proton and the neutron. Putting these values in the expression, we get,

\[Q=(238.050770\,u-234.043583\,u-4.002603\,u)\times931.414{\mathrm{MeV}\over \mathrm{u}}\]

Here we’ve used the fact that each atomic mass unit is worth $931.414$ MeV of energy. (Since we’re using this direct conversion from mass to energy, we needn’t worry about the $c^2$ factor.) Simplifying further we obtain,

\[Q=4.2696\, \mathrm{MeV}\]

This is not all the energy the alpha particle ejects out with, since as we mentioned above, part of the energy is used up in the daughter nucleus recoiling backwards. The precise kinetic energy with which the alpha particle escapes the parent nucleus is calculated according to the Eq($\ref{eq: Kinetic energy}$).

\begin{align*}

K_{\alpha}&=\left({234.043583\over 4.002603+234.043583}\right)4.2696\\

&=0.9831\times 4.2696\\

&=4.1974 \,\mathrm{MeV}\\

&\approx 4.2 \,\mathrm{MeV}

\end{align*}

This is a relatively lower energy when compared with the energies involved with other radioactive decays such as the beta and the gamma.

When understanding the process of alpha decay, one of the first questions that might crop up in the mind is if the stability of the nucleus is the prime goal to be acheived by losing some nucelons, why only alpha particles are ejected, why not a single proton or some other combinations of protons and neutrons?

The answer to this lies in the value of the kinetic energy we obtained above. In order to escape the parent nucleus, the ejecting particle (essentially a chunck of the parent) must have enough kinetic energy to overcome the binding energies that pull hard to keep the nucleus together. It turns out that the alpha particle has that necessary energy and moreover the least when compared with the kinetic energies involved if other combinations of nucleons were ejected. You may repeat the above procedures and verify yourselves that the kinetic energies would be larger if a single proton were to eject, or any other combination of nucleons such as ${}^3_2\mathrm{He}$. Nature not just restores balance but does so in the most efficient manner.

The question still remains, what happens of the alpha particle after the decay? The answer to that depends on where the alpha particle is headed to. If it simply flies out into open air, it will collide with the air molecules and at each collision, it loses its energy, hence slowing down. This lost energy is mostly abosrbed by the very light electrons, either through electric attarctions (remember the alpha partcile is positvely charged, capable of tugging at the negatively charged electrons) or by rightaway knocking them off the atom and freeing them from the atomic bind. Thus the alpha particle is capable of producing pairs of oppositely charged ions as it traverses through matter – the negatively charged free electron and the positively charged atom (that lost the electron).

Due to its large mass compared to the electrons, the alpha particle literally ploughs through them. While there are many electrons that are set free by these collisions, there are many others that couldn’t receive enough energy from the collisions to escape the atom but gained some energy (and got excited to a higher energy level). Such electrons ultimately fall back to their original state by giving back the excess gained energy, many times in the form of visible light. This is just the phenomenon of phosporoscence that we discussed in the beginning. It enables objects to glow in the dark. Therefore it must be made clear that radioacitive sources themselves do not glow in the dark, as most science fiction tales depict, but when brought near certain (phosphoroscent) materials, they can cause those materials to glow.

The law of energy conservation that gave the alpha particles the energy after they left the parent nucleus also takes back the energy from them in the form of a loss of energy due to the collisions, and the alpha particle ultimately comes to a halt, having lost all its energy.

Collisions with the atomic nuclie are possible but quite rare due to the tiny size of the nuclei. The nucleus is at least $100,000$ times smaller than the atom (this is similar to the comparison of the size of a pingpong ball with 20 times the size of a standard football field). Hence it is very likely that the alpha particles miss the nuclei. However improbable, such collisions are still possible and when they do happen, the alpha particles get deflected wildly. This is because the nucleus is positively charged and so is the alpha particle. But unlike the electron, the atomic nucleus is extremely dense and hence massive. A head-on collision with the nucleus can even send back the alpha particle along the path that it came (a 180 degree bounce). In fact it was such observations that prompted Lord Rutherford during the ealry 20th C to speculate the structure of the atom.

As for the possible dangers of alpha radiations, you don’t have to worry if the only way alpha rays can try enter your body is through your skin. Due to the low energies, the alpha particles cannot cross even the outer layer of the skin. You still ought to keep away from any radioactivity, for though alpha decays seem harmless (externally), most alpha decays are followed by a gamma decay (bursts of high energy photons) that is extremely penetrative and can cause serious damage to your body.

Also if the radiation has contaminated the air around you and you inhaled this air taking radioactive atoms inside your body, it can mean serious trouble. The atoms will continue emitting the alpha particles (with energies that are sufficient to damage your internal organs) for a very long time. The duration of the decays can be especially prolonged due to the fact that the residue of an alpha decay is most often radioactive too and will again begin its own emissions.

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