Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the following problem:

The nuclei of Am-241 decay by the emission of $\alpha$-particles with a kinetic speed of $8.8 \cdot 10^{-19} J$. In a certain source of Am-241 there are $ 4.0 \cdot 10^3$ decaying nuclei per second. Caculate the mass this source loses per year.

So what the correction sheet tells me is that you basically just have to find the mass of an $\alpha$-particle in $kg$ and then just multiply that number until you get the amount of decaying $\alpha$-particles in a year.

However, I think this is wrong, since the $8.8 \cdot 10^{-19} J$ isn't accounted for. I'd say you have to take the mass of the $\alpha$-particle, then add to that the $8.8 \cdot 10^{-19} J$ after you've converted it to $kg$ by using $E = mc^2$ and THEN multiplying that number in order to get the amount of decaying $\alpha$-particles in a year.

Which is correct, or perhaps better: which is less wrong?

share|cite|improve this question

Work it out both ways and compare the difference in mass, you should get it being zero to all intents an purposes

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.