Alpha Ray Interactions with Au and Be: Unraveling the Neutron Tear Out Mystery

[onurbeyaz]

Why do alpha ray tear out neutron in Be, but don't do such thing in Au?

 

[Simon Bridge]

It is difficult to know how to answer that other than to say "because it can" but using bigger words.

1. we talk about alpha particles rather than rays
2. "tearing out" is not a good way to think of what happens
3. some isotopes of gold do undergo alpha-decay

If you think of a nucleus as a kind of liquid that is constantly shaking, then some nuclei shake hard enough that droplets fly off, some don't, it depends on how hard the nuclei is held together, which depends on the balance between protons and neutrons - the strong force vs the coulomb force.

 

[onurbeyaz]

It is difficult to know how to answer that other than to say "because it can" but using bigger words.

1. we talk about alpha particles rather than rays
2. "tearing out" is not a good way to think of what happens
3. some isotopes of gold do undergo alpha-decay

If you think of a nucleus as a kind of liquid that is constantly shaking, then some nuclei shake hard enough that droplets fly off, some don't, it depends on how hard the nuclei is held together, which depends on the balance between protons and neutrons - the strong force vs the coulomb force.

Thanks for the answer, so can we say that it is because the strenght of coulomb force in this situation is usually higher than atomic bounds in Be, but lower than atomic bounds in Au(I even don't know if bounds are more strenght in Au than Be)

 

[e.bar.goum]

It's nothing about atomic bonds. This is pure nuclear physics! But you're correct to invoke the coloumb force.

Beryllium-9 is a "fragile" or "weakly bound" nucleus, and it's quite easy to liberate a neutron in a direct breakup process:

9Be + a -> 3a + n.

The threshold for this is 1.57 MeV, so it's not hard to liberate a neutron. This is why a common laboratory source of neutrons is an Am-Be source - the Am decays giving ~5.1 MeV alpha particle, which breaks up the Be, giving a neutron.

On the other hand, if you consider the knockout reaction

196Au+4He+n, you need at least 8.07 MeV

The same cannot be said for 197Au.

This is not even considering the different charges of 197Au and 9Be - the extra 75 protons will make it much harder for the alpha particle to come in close enough to knock out a neutron from gold.

 

[onurbeyaz]

It's nothing about atomic bonds. This is pure nuclear physics! But you're correct to invoke the coloumb force.

Beryllium-9 is a "fragile" or "weakly bound" nucleus, and it's quite easy to liberate a neutron in a direct breakup process:

9Be + a -> 3a + n.

The threshold for this is 1.57 MeV, so it's not hard to liberate a neutron. This is why a common laboratory source of neutrons is an Am-Be source - the Am decays giving ~5.1 MeV alpha particle, which breaks up the Be, giving a neutron.

On the other hand, if you consider the knockout reaction

196Au+4He+n, you need at least 8.07 MeV

The same cannot be said for 197Au.

This is not even considering the different charges of 197Au and 9Be - the extra 75 protons will make it much harder for the alpha particle to come in close enough to knock out a neutron from gold.

Thanks for the answer. I don't know any of the nuclear physics so this is much more than I expected. Is this why Rutherford used Gold in his experiment? What will happen if we use any other material such as any other metal like iron, or non metal like carbon

 

[e.bar.goum]

Thanks for the answer. I don't know any of the nuclear physics so this is much more than I expected. Is this why Rutherford used Gold in his experiment? What will happen if we use any other material such as any other metal like iron, or non metal like carbon

In the Geiger-Marsden experiment (aka the Rutherford experiment, but Geiger and Marsden did the work) you use gold because it's heavy and it's pure. The fact it has a lot of protons means you see the effect of elastic scattering so much more. You could use pretty much any element, but you're best off using something like gold or lead, since you will get more back-angle scattering. The chemistry does not matter at all - you can use a metal or a non-metal.

The ability to free a neutron has nothing at all to do with this experiment, as you're only investigating the Coulomb part of the nuclear potential - although if you used a weakly bound nucleus you would start to see breakup which would make your results a bit confusing.

 

[Simon Bridge]

@e.bar: isn't onerbeyaz asking about alpha decay?

Certainly the Rutherford exp could be done with any atoms... and any charged particle. Au and alphas were used to help control variables and they were handy.

Quite a lot of experiments end up using particular materials over others simply because the experimenter happened to have some in the lab at the time.

 

[e.bar.goum]

I interpreted "Why do alpha ray tear out neutron in Be" as an alpha particle reacting with Be to produce a neutron rather than alpha decay, as both 9Be and 197Au are stable to alpha decay. Using alpha+9Be as a neutron source is a very common trick.

onerbeyaz was asking about the rutherford experiment previously, which I assume they are referring to now?

Yeah, that's definitely the case with gold in experiments these days, and I assume it was the same with Geiger and Marsden - 197Au is the only stable isotope of gold, so if you want an isotopically pure sample of anything on the cheap, you go with gold (other experimental considerations not withstanding of course). It's also very easy to work - you can have a self-supporting target virtually as thin as you want. Gold is very commonly used as a calibration target for elastic scattering.