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 $\Delta m$?

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?
$\Delta m=Zm_p+Nm_n-m$ 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? Apr 22, 2013 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)... Apr 22, 2013 at 11:41