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

Step-by-step explanation:

The rate of radio active decay is given as:

`A(t) = A(0)e^(-0.693t/t1/2) \\\\= 205 g * e^{(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)=(0.693)/(k)`

Putting value of
`t_(1/2)` in this formula, we get the rate constant.

`103.5* 10^6=(0.693)/(k)`

`k=6.6* 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=(2.303)/(t)\log(a)/(a-x)`

where,

k = rate constant =
`6.6* 10^(-9)year^(-1)`

t = time taken for decay process =
`1.035* 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* 10^(-9)=(2.303)/(1.035* 10^(9))\log(205)/(a-x)`

`a-x=0.200g`

Therefore, the amount left of the samarium-146 is, 0.200 g