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 :
[tex]t_{1/2}=\frac{0.693}{k}[/tex]
Putting value of [tex]t_{1/2}[/tex] in this formula, we get the rate constant.
[tex]103.5\times 10^6=\frac{0.693}{k}[/tex]
[tex]k=6.6\times 10^{-9}year^{-1}[/tex]
Now we have to calculate the original amount of samarium-146.
The expression for rate law for first order kinetics is given by :
[tex]k=\frac{2.303}{t}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]6.6\times 10^{-9}year^{-1}[/tex]
t = time taken for decay process = [tex]1.035\times 10^{9}years[/tex]
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.
[tex]6.6\times 10^{-9}=\frac{2.303}{1.035\times 10^{9}}\log\frac{205}{a-x}[/tex]
[tex]a-x=0.200g[/tex]
Therefore, the amount left of the samarium-146 is, 0.200 g
Ans: 0.200 g
Given:
Half life of Sm-146 = t1/2 = 103.5 million years
Time period, t = 1.035 billion years = 1035 million years
Original mass of sample, [A]₀ = 205 g
To determine:
Amount of sample after t = 1035 million years
Explanation:
The rate of radio active decay is given as:
[tex]A(t) = A(0)e^{-0.693t/t1/2} \\\\= 205 g * e^{\frac{0.693*1035}{103.5} } \\\\= 0.200 g[/tex]
