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Forum Rules: Read This Before Posting Topic: Radioactive decay energy 2  (Read 2364 times)

0 Members and 1 Guest are viewing this topic. Rutherford « on: December 10, 2012, 08:22:24 AM »
When I saw the post below I remembered of one thing I wasn't sure about decay energy. The formula is Q = minitialc2 - mfinalc2.
I found here http://web.utk.edu/~cnattras/Phys250Fall2012/modules/module%205/nuclear_decay.htm when Q<0, energy is spent and when Q>0 energy is released which is the opposite of thermodynamic laws.
Is the formula in the link wrong or is it a contradiction? If it is a contradiction I could write the formula this way Q = mfinalc2 - minitialc2, wouldn't it then be okay with thermodynamic? Bublik

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• Mole Snacks: +5/-1 « Reply #1 on: December 10, 2012, 09:11:26 PM »
Yeah, if we're being very technical, you're right. But the sample problems on that link asked to calculate the "disintegration energy" (which can be interpreted as a magnitude so really it doesn't matter whether it's positive or negative).

The formula Q = mfinalc2 - minitialc2 is okay with thermodynamics when calculating change in energy (which of course is always negative, exothermic, for decay reactions).

But once again, the questions on that link are pretty much asking "how much energy is released?" in which you would just reply with a magnitude, since negative is already implied. Rutherford « Reply #2 on: December 11, 2012, 07:24:19 AM »
If you read the first problem in alpha decay section, you would see that the sign before the number is minus and then they wrote:
''The energy released is negative, we have to add energy to make this reaction possible.''
This confused me.
You said that the other formula is okay with thermodynamics and I will probably stick to it, so there won't be any confusion.

One more question. Is it an approximation to use the mass of He for an alpha particle, as the alpha particle doesn't have electrons? Is it okay? Rutherford 