Question

Strontium−90 is one of the products of the fission of uranium−235. This strontium isotope is radioactive, with a half-life of 28.1 yr. Calculate how long (in yr) it will take for 1.00 g of the isotope to be reduced to 0.600 g by decay.

Answer 1

Answer : The time passed in years is 20.7 years.

Explanation :

Half-life = 28.1 years

First we have to calculate the rate constant, we use the formula :

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

`k=\frac{0.693}{28.1\text{ years}}`

`k=2.47* 10^(-2)\text{ years}^(-1)`

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

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

where,

k = rate constant =
`2.47* 10^(-2)\text{ years}^(-1)`

t = time passed by the sample = ?

a = initial amount of the reactant = 1.00 g

a - x = amount left after decay process = 0.600 g

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

`t=(2.303)/(2.47* 10^(-2))\log(1.00)/(0.600)`

`t=20.7\text{ years}`

Therefore, the time passed in years is 20.7 years.