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$\require{mhchem}$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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The actual mass after the nucleus has formed is measured using mass spectrometers. The assumed mass is predicted using the number of protons and neutrons and finding their total mass. In reality, the actual mass is always less than the assumed mass. That's where mass defect comes in...

$\Delta m=Zm_p+Nm_n-m$ 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 $Q=\Delta m c^2$

You should take into account of all the products when calculating the energy released by the nuclear reaction. That's all I wanted to say...

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    $\begingroup$ 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? $\endgroup$
    – Jiew Meng
    Apr 22, 2013 at 3:50
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    $\begingroup$ 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)... $\endgroup$ Apr 22, 2013 at 11:41

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