During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1μg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.
k = \(\frac{0.693}{t_{\frac{a}{2}}}\) = \(\frac{0.693}{28.1}\)year-1
Here,
It is know that,
t = \(\frac{2.303}{k}\)log \(\frac{[R]_0}{[R]}\)
⇒ 10 = \(\frac{2.303}{\frac{0.693}{28.1}}\)log \(\frac{1}{[R]}\)
⇒ 10 = \(\frac{2.303}{\frac{0.693}{28.1}}\)(-log [R])
⇒ log [R] = -\(\frac{10 \times 0.693}{2.303 \times 28.1}\)
⇒ [R] = antilog (-0.1071)
= antilog \((\bar{1}.8929)\)
= 0.7814 μg
Therefore, 0.7814 μg of 90Sr will remain after 10 years.
Again,
t = \(\frac{2.303}{k}\)log \(\frac{[R]_0}{[R]}\)
⇒ 60 = \(\frac{2.303}{\frac{0.693}{28.1}}\)log \(\frac{1}{[R]}\)
⇒ log [R] = -\(\frac{60 \times 0.693}{2.303 \times 28.1}\)
⇒ [R] = antilog (-0.6425)
= antilog \((\bar{1}.3575)\)
= 0.2278 μg
Therefore, 0.2278 μg of 90Sr will remain after 60 years.