# Computing $\Delta m$ in $\beta^-$ decay

$\require{mhchem}$What is the energy $Q$ released when $\ce{^131_53I}$ decays and $\ce{^131_54 Xe}$ is formed? The atomic mass of $\ce{I}$ is $130.906118~u$ and the atomic mass of $\ce{Xe}$ is $130.90508~u$.

To compute $\Delta m$, I thought of $130.90508~u - 130.906118~u = 0.298962~u$ but answer is $0.001038~u$ way off. How do I actually compute $\Delta m$?

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I'm having a little trouble reading your question. It sort of looks like you just made a math mistake or typo: $130.906118 u - 130.90508u = 0.001038$, but you seem to indicate that it equals $.298962 u$. Maybe I'm just misreading?

But in anycase, you have the right idea. You don't have to worry about the mass of the electron, as the numbers are atomic masses, and thus include the electrons already.

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$\Delta m=Zm_p+Nm_n-m$ is the mass defect of a nucleus with $Z$ protons and $N$ neutrons (but, that's technical). In case of nuclear reactions, it can be simply calculated by using $\Delta m=m_{reactants}-m_{products}$ and so, the energy released would be $Q=\Delta m c^2$
In $\Delta m=Zm_p+Nm_n-m$, whats $m$? Also the number of neutrons and protons will be for which element? I or Xe? –  Jiew Meng Apr 22 at 3:50
Hi @JiewMeng: The $m$ is the predicted mass (from experiments). The remaining things in the equation contribute to the calculated mass. By that expression, I mentioned that the $\Delta m$ can be calculated for individual elements. Doing the same for nuclear reactions makes somewhat complicated (good for approximation though). So, I suggest the use of the second one. What I really mention is that you didn't take $m_e$ into account. In order for the conservation of momentum, you've to balance the products with the reactants (both mass & energy)... –  Waffle's Crazy Peanut Apr 22 at 11:41