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Question: During a nuclear explosion, one of the products is \(^{90}Sr\) with a half-life of \(28.1\) years. I...

During a nuclear explosion, one of the products is 90Sr^{90}Sr with a half-life of 28.128.1 years. If 1μg1\mu g of 90Sr^{90}Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 1010 years and 6060 years if it is not lost metabolically.

Explanation

Solution

Radioactive decay is basically a process in which the unstable nucleus loses its energy in the form of radiation and that substance that contains such an unstable nucleus is known as radioactive substances. So, we need to find out the amount of 90Sr^{90}Srleft that we will find out by manipulating a formula.

Complete step by step answer:
We know that all the radioactive decay has first order reactions. We will use a formula to find out-
k=2.303tlog⁡N0Nk = \dfrac{{2.303}}{t}\log \dfrac{{{N_0}}}{N}, where kkis radioactive decay constant
N0{N_0} is initial activity
NN is final activity
tt if half life
Now, first we need to find out the half-life, which we will find out using:
k=0.693t12=0.69328.1=0.0247year−1k = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}} = \dfrac{{0.693}}{{28.1}} = 0.0247yea{r^{ - 1}}
i) 10years10years
k=2.303tlog⁡N0Nk = \dfrac{{2.303}}{t}\log \dfrac{{{N_0}}}{N}, using this formula we will find out the amount left.
We have,
t=10yearst = 10years, N0=1microgram=1×10−6g{N_0} = 1microgram = 1 \times {10^{ - 6}}g,
Now, after putting these values in the above given formula,
10=2.3030.0247log⁡1×10−6N10 = \dfrac{{2.303}}{{0.0247}}\log \dfrac{{1 \times {{10}^{ - 6}}}}{N}
log⁡1×10−6N=10×0.02472.303=0.1072\log \dfrac{{1 \times {{10}^{ - 6}}}}{N} = \dfrac{{10 \times 0.0247}}{{2.303}} = 0.1072
1×10−6N=Antilog⁡0.1072\dfrac{{1 \times {{10}^{ - 6}}}}{N} = Anti\log 0.1072
=1.1280= 1.1280
Now, we will take NN to the other side,
N=1×10−61.1280=0.7842μgN = \dfrac{{1 \times {{10}^{ - 6}}}}{{1.1280}} = 0.7842\mu g
So, after 1010 years, 0.7842μg0.7842\mu g will be left undecayed.
ii) 60years60years
Now, we will solve for 60years60years in the same way we did for 10years10years
k=2.303tlog⁡N0Nk = \dfrac{{2.303}}{t}\log \dfrac{{{N_0}}}{N}
We have t=60yearst = 60years, N0=1microgram=1×10−6g{N_0} = 1microgram = 1 \times {10^{ - 6}}g
After putting these values in the formula, we get:
60=2.3030.0247log⁡1×10−6N60 = \dfrac{{2.303}}{{0.0247}}\log \dfrac{{1 \times {{10}^{ - 6}}}}{N}
1×10−6N=Antilog⁡0.6453=4.400\dfrac{{1 \times {{10}^{ - 6}}}}{N} = Anti\log 0.6453 = 4.400
After taking N to the other side, we get:
N=1×10−64.400=0.227μgN = \dfrac{{1 \times {{10}^{ - 6}}}}{{4.400}} = 0.227\mu g
Therefore, 0.227μg0.227\mu g will be left undecayed after 60years60years

Note:
In radioactive substances, when there is a spontaneous breakdown of its atomic nucleus then that results in the emission of radiation from that nucleus and we call that as Radioactive decay.