Question

Thorium 234 is a radioactive substance that decays at a rate proportional to the amount present. 1 gram of this material is reduced to 0.8 grams in one week

(a) Find an expression that expresses the amount of Thorium 234 present at any time

(b) Find half life of Thorium 234

(c) Find amount of Thorium 234 present after 10 weeks

Answer 1

Answer: a)
`t=(2.303)/(k)\log(a)/(a-x)`

b) 3.15 weeks.

c) 0.11 grams

Explanation:

a) radioactive decay follows first order kinetics and thus Expression for rate law for first order kinetics is given by:

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

where,

k = rate constant = ?

t = time taken for decomposition = 1 week

a = initial amount of the reactant = 1 g

a - x = amount left after decay process = 0.8 g

Now put all the given values in above equation, we get

`k=(2.303)/(1)\log(1)/(0.8)`

`k=0.22weeks^(-1)`

2) To calculate the half life, we use the formula :

`t_{(1)/(2)=(0.693)/(k)`

`t_{(1)/(2)=(0.693)/(0.22)=3.15weeks`

Thus half life of Thorium 234 is 3.15 weeks.

3) amount of Thorium 234 present after 10 weeks:

`10=(2.303)/(0.22)\log(1)/(a-x)`

`(a-x)=0.11g`

Thus amount of Thorium 234 present after 10 weeks is 0.11 grams