Question

Samarium-146 has a half-life of 103.5 million years. After 1.035 billion years, how much samarium-146 will remain from a 205-g sample?
0.200 g
0.400 g
20.5 g
103 g

Answer 1

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:

$A(t) = A(0)e^{-0.693t/t1/2} \\\\= 205 g * e^{\frac{0.693*1035}{103.5} } \\\\= 0.200 g$

Answer 2

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