Question

sodium-22 undergoes positron emission as shown:
22/11 na → 22/10 ne + 0/1 β
if this decay has a half - life of 2.60 years, what mass of 72.5 g sodium - 22 will remain after 15.6 years?
——g
do not include units in your answer. your answer should have three significant digits.

Explanation:

Step1: Calculate the number of half - lives

The formula for the number of half - lives \(n=\frac{t}{T}\), where \(t = 15.6\) years (time elapsed) and \(T=2.60\) years (half - life).
\(n=\frac{15.6}{2.60}=6\)

Step2: Use the radioactive decay formula

The radioactive decay formula is \(N = N_0\times(\frac{1}{2})^n\), where \(N_0 = 72.5\) g (initial mass) and \(n = 6\) (number of half - lives).
\(N=72.5\times(\frac{1}{2})^6\)
First, calculate \((\frac{1}{2})^6=\frac{1}{64}\)
Then \(N = 72.5\times\frac{1}{64}\)
\(N=\frac{72.5}{64}=1.1328125\approx1.13\) g (rounded to three significant digits)

Answer:

\(1.13\)