Answer : The correct option is, 0.200 g
Solution :
As we know that the radioactive decays follow first order kinetics.
First we have to calculate the rate constant of a samarium-146.
Formula used :
\(t_{1/2}=\frac{0.693}{k}\)
Putting value of \(t_{1/2}\) in this formula, we get the rate constant.
\(103.5\times 10^6=\frac{0.693}{k}\)
\(k=6.6\times 10^{-9}year^{-1}\)
Now we have to calculate the original amount of samarium-146.
The expression for rate law for first order kinetics is given by :
\(k=\frac{2.303}{t}\log\frac{a}{a-x}\)
where,
k = rate constant = \(6.6\times 10^{-9}year^{-1}\)
t = time taken for decay process = \(1.035\times 10^{9}years\)
a = initial amount of the samarium-146 = 205 g
a - x = amount left after decay process = ?
Putting values in above equation, we get the value of initial amount of samarium-146.
\(6.6\times 10^{-9}=\frac{2.303}{1.035\times 10^{9}}\log\frac{205}{a-x}\)
\(a-x=0.200g\)
Therefore, the amount left of the samarium-146 is, 0.200 g
