The halflife of cesium137 is 30 yearsSuppose we have a 10g s

The half-life of cesium-137 is 30 years.Suppose we have a 10-g sample. After how long will only 3 g of the sample remain

Solution

We have to use the following formula

A=A₀*(1/2)^t/h

A is the final amount which is 3 gm.

A₀ is the initial amount which is 10gm

h is the half life which is 30 yrs

On using the above formula we get

3=10*(1/2)^t/30

.3= (1/2)^t/30

Taking log on both sides

Log .3= (t/30) log (1/2)

t= 30((log .3)/ log(.5))

t= 52 yrs

Therefore after 52 yrs only 3g of the sample remain