# Why do products of nuclear decay have a lower mass than the original nucleus, when the sum of the mass of its nucleons is larger? [duplicate]

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I've just started covering the topic of binding energy in Year 13 at school (final year before University). The definition we've been given of binding energy is that it is the work done when separating a nucleus into its constituent nucleons. Alright so far.

We've also covered nuclear decays, and an example of that we've been given is where the mass of the Thorium isotope and the alpha particle (constituent parts) is less than the mass of the Uranium isotope (I've checked this - it's true). We were told this was due to the combined binding energy of the products being less than the binding energy of the uranium isotope.

However, this seems to contradict examples I've found on the internet, such as the formation of an alpha particle from 2 protons and 2 neutrons (see here), where the mass of the constituent parts is higher - energy must remain constant, and the alpha particle has a higher binding energy, so the mass-energy must necessarily be lower. This makes sense.

If someone is able to give a clear explanation of why both of these things are true, I'd be very grateful!

Thanks

## marked as duplicate by John Rennie, Daniel Griscom, Gert, user36790, ACuriousMind♦Jan 22 '16 at 2:05

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• I agree with @JohnRennie - remember that $E = mc^{2}$ - nuclear decays go to a lower overall energy state, so that energy goes away and the mass must be lower. – Jon Custer Jan 21 '16 at 17:50
• I've already read that post actually - I understand the explanation of why a greater binding energy decreases the mass of the isotope compared to its constituent nucleons. However, I don't understand why the combination of the thorium isotope and alpha particle (which have a combined binding energy lower than that of the uranium) would be lower in mass. – Swin Jan 21 '16 at 17:51
• If I've said it once I said it a mill....well, ten or twenty .... times: a bound nucleon is not the same as a free nucleon. Nor do you expect conservation of mass in nuclear systems. – dmckee Jan 21 '16 at 22:26

## 1 Answer

The mass defect and the binding energy are not linear functions of the number of nucleons. They increase until the iron and then decrease.

See the Figure 31-5 of The Mass Defect of the Nucleus and Nuclear Binding Energy The thorium Mass Defect for one Average Nucleon ( MDAN ) is higher than an uranium MDAN.

Uranium decays in thorium and alpha particle and produces energy

But, in contrast, one hydrogene MDAN is lower than 1 helium MDAN ( the curve ).

1 alpha particle needs energy to decay in Hydrogene and free neutrons or deuterium .

Then all is for the best.

For the fun, notice the Oxygene and Carbon catastroph :)

• Thanks for this! I didn't realise that the Thorium and alpha particle have a larger combined binding energy than the uranium! Also, thanks for actually answering my question; people who simply marked my post as a duplicate obviously didn't read it properly – Swin Jan 22 '16 at 10:18