New answers tagged radioactivity
1
The mass of the original polonium atom is 209.9828737(13)AU, while the mass of the lead atom is 205.9744653(13)AU and the mass of the helium is 4.00260325415(6)AU. The mass deficit gives you the amount of energy released.
3
The core reason is that the chance for a given atom to decay in the next $n$ seconds is always the same.
You can get a more intuitive feel for this by considering a simple game with dice.
Start with some initial population of ordinary dice (I'll assume we're using the basic six-sided (cubical) ones, but you table top RPG fiends can use polyhedral ones if ...
2
So here's what I did, I discretized the problem and deduced $N_x$ and $N_y$ at some $t_n$. These happened to include sums that could easily be turned into integrals. Namely:
$$N_x(t_n)=N_0e^{-\lambda t_n}+\Lambda\sum_{i=0}^n\Delta t_ie^{-\lambda(t_n-t_i)}$$
$$N_y(t_n)=N_0(1-e^{-\lambda t_n})+\Lambda\sum_{i=0}^n\Delta t_i(1-e^{-\lambda(t_n-t_i)})$$
As a ...
2
The first of your equations is correct. You can see this in two ways. First, just look at the dimensions. In general, the argument of a logarithm should be dimensionless; only your first option is. Second, and maybe more convincingly, look at what you get when you take $\Lambda \to0$. You should be able to reproduce the standard decay equation:
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