During the nuclear explosion, one of the products is $^{90} Sr$ with half life of 6.93 years. If 1 $μ g$ of $^{90} Sr$ was absorbed in the bones of a newly born baby in place of $C a$ , how much time, in years, is required to reduce it by 90% if it is not lost metabolically?

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Solution

All nuclear processes follow first order kinetics, and hence
$t_{1 / 2} = \frac{0.693}{λ}$
$λ = 0.1 (year)^{- 1}$
$t = \frac{2.303}{λ} log (\frac{a_{0}}{a_{t}})$
$t = \frac{2.303}{0.1} log (\frac{a_{0}}{0.1 a_{0}})$
On solving, $t = 23.03$ years

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During the nuclear explosion, one of the products is 90Sr with half life of 6.93 years. If 1 μg of 90Sr was absorbed in the bones of a newly born baby in place of Ca, how much time, in years, is required to reduce it by 90% if it is not lost metabolically?

All nuclear processes follow first order kinetics, and hence t1/2=0.693λ λ=0.1(year)−1 t=2.303λlog(a0at) t=2.3030.1log(a00.1a0) On solving, t=23.03 years

During the nuclear explosion, one of the products is $^{90} Sr$ with half life of 6.93 years. If 1 $μ g$ of $^{90} Sr$ was absorbed in the bones of a newly born baby in place of $C a$ , how much time, in years, is required to reduce it by 90% if it is not lost metabolically?

All nuclear processes follow first order kinetics, and hence
$t_{1 / 2} = \frac{0.693}{λ}$
$λ = 0.1 (year)^{- 1}$
$t = \frac{2.303}{λ} log (\frac{a_{0}}{a_{t}})$
$t = \frac{2.303}{0.1} log (\frac{a_{0}}{0.1 a_{0}})$
On solving, $t = 23.03$ years