Question

During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1μg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.

Answer 1

Dear Aravind,

◆ Answer -

After 10 years, A1 = 0.7808 μg

After 60 years, A2 = 0.2266 μg

● Explanation -

# Given -

t½ = 28.1 yrs

A = 1 μg

t1 = 10 yrs

t2 = 60 yrs

# Solution -

Radioactive decay constant λ is calculated as -

λ = 0.693 / t½

λ = 0.693 / 28.01

λ = 0.02474 /yrs

Remaining amount of 90Sr after 10 years is -

A1 = A e^(-λt2)

A1 = 1 × e^(-0.02474×10)

A1 = e^(-0.2474)

A1 = 0.7808 μg

Remaining amount of 90Sr after 60 years is -

A2 = A e^(-λt2)

A2 = 1 × e^(-0.02474×60)

A2 = e^(-1.4844)

A2 = 0.2266 μg

Hope that is helpful...

Answer 2

Answer:

` t= 2.303/k log  ([R]_0)/([R])`

`=> 60 = 2.303/(0.693/28.1) log  1/[R]`

`=>log[R] = -(60xx0.693)/(2.303xx28.1)`

=>[R] = antilog (-0.6425)

=antilog(`bar"1"`.3575)

= 0.2278 `mug`

Therefore, 0.2278 μg of 90Sr will remain after 60 years