Question

Suppose a laboratory has a 38 g sample of polonium-210. the half-life of polonium-210 is about 138 days.

a. how many half-lives of polonium-210 occur in 1104 days?
b. how much polonium is in the sample 1104 days later?

Answer 1

a. 1104 / 138 = 8 half lives

b after 1104 days amount of polonium in sample = 38 (1/2)^8 = 0.184 g to nearest thousandth of a gram

Answer 2

(a) The number of half-lives of polonium-210 are, 8

(b) The amount left in the sample after 1104 days will be, 0.391 grams.

Explanation :

First we have to determine the amount left in the sample after 1104 days.

This is a type of radioactive decay and all radioactive decays follow first order kinetics.

To calculate the rate constant, we use the formula :

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

`k=\frac{0.693}{138\text{ days}}`

`k=5.022* 10^(-3)\text{ days}^(-1)`

Now we have to calculate the amount left.

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 decay process

a = initial amount or moles of the reactant

a - x = amount or moles left after decay process

Putting values in above equation, we get:

`5.022* 10^(-3)=\frac{2.303}{138\text{ days}}\log(100g)/(a-x)`

`a-x=0.391g`

The amount left in the sample after 1104 days will be, 0.391 grams.

Now we have to calculate the number of half-lives of polonium-210.

`a=(a_o)/(2^n)`

where,

a = amount of reactant left after n-half lives = 0.391 g

`a_o` = Initial amount of the reactant = 100 g

n = number of half lives = ?

Putting values in above equation, we get:

`0.391=(100)/(2^n)`

`256=2^n`

`2^8=2^n`

`n=8`

Therefore, the number of half-lives of polonium-210 are, 8