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Xelga [282]
3 years ago
11

Based on nuclear stability, what is the symbol for the most likely product nuclide when iodine-123 undergoes decay?

Chemistry
1 answer:
MissTica3 years ago
6 0
Answer is: Te (tellurium).
During electron capture, iodine-123 f<span>orm the nearly-stable nuclide tellurium-123.
</span>In beta plus decay (atomic number Z is decreased by one), a proton is converted to a neutron and positron and an electron neutrino, so mass <span>number does not change.
</span>Iodine-123 is a radioactive isotope of iodine used in nuclear medicine imaging.<span>
</span>
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If 27.3% of a sample of silver-112 decays in 1.52 hours, what is the half-life (in hours to 3 decimal places)?
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<u>Answer:</u> The half life of the sample of silver-112 is 3.303 hours.

<u>Explanation:</u>

All radioactive decay processes undergoes first order reaction.

To calculate the rate constant for first order reaction, we use the integrated rate law equation for first order, which is:

k=\frac{2.303}{t}\log \frac{[A_o]}{[A]}

where,

k = rate constant = ?

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[A_o] = Initial concentration of reactant = 100 g

[A] = Concentration of reactant left after time 't' = [100 - 27.3] = 72.7 g

Putting values in above equation, we get:

k=\frac{2.303}{1.52hrs}\log \frac{100}{72.7}\\\\k= 0.2098hr^{-1}

To calculate the half life period of first order reaction, we use the equation:

t_{1/2}=\frac{0.693}{k}

where,

t_{1/2} = half life period of first order reaction = ?

k = rate constant = 0.2098hr^{-1}

Putting values in above equation, we get:

t_{1/2}=\frac{0.693}{0.2098hr^{-1}}\\\\t_{1/2}=3.303hrs

Hence, the half life of the sample of silver-112 is 3.303 hours.

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