Question

A radioactive isotope has a half-life of Tyears. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value?

Answer 1

Dear Student,

◆ Answer -

a) t1 = 5T

b) t2 = 6.646T

● Explaination -

decay constant λ is calculated as -

λ = 0.693/t½

λ = 0.693/T

Time required for radioisotope to reduce from Ao to A is -

t = 2.303/λ × log(Ao/A)

t = 2.303/(0.693/T) × log(Ao/A)

t = 3.323T × log(Ao/A)

a) For A1 = Ao × 3.125/100 = Ao/32,

t1 = 3.323T × log(Ao/A1)

t1 = 3.323T × log(32)

t1 = 3.323T × 5 × log(2)

t1 = 3.323T × 5 × 0.301

t1 = 5T

Therefore, the isotope will take about 5T years to reduce to 3.125% of its original value.

b) For A2 = Ao × 1/100 = Ao/100,

t2 = 3.323T × log(Ao/A2)

t2 = 3.323T × log(100)

t2 = 3.323T × 2 × log(10)

t2 = 3.323T × 2 × 1

t2 = 6.646T

Therefore, the isotope will take about 6.646T years to reduce to 1% of its original value.

Hope this helps you...