Question

A sample contains radioactive atoms of two types, A and B. Initially there are five times as many A atoms as there are B atoms. Two hours later, the numbers of the two atoms are equal. The half-life of A is 0.77 hours. What is the half-life of B in hours?

Answer 1

Step-by-step explanation:

Initially no of atoms of A = N₀(A)

Initially no of atoms of B = N₀(B)

5 X N₀(A) = N₀(B)

N = N₀
`e^(-\lambda t)`

N is no of atoms after time t , λ is decay constant and t is time .

For A

N(A) = N(A)₀
`e^(-\lambda_1 t)`

For B

N(B) = N(B)₀
`e^(-\lambda_2 t)`

N(A) = N(B) , for t = 2 h

N(A)₀
`e^(-\lambda_1 t)` = N(B)₀
`e^(-\lambda_2 t)`

N(A)₀
`e^(-\lambda_1 t)` = 5 x N₀(A)
`e^(-\lambda_2 t)`

`e^(-\lambda_1 t)` = 5
`e^(-\lambda_2 t)`

`e^(\lambda_2 t)` = 5
`e^(\lambda_1 t)`

half life = .693 / λ

For A

.77 = .693 / λ₁

λ₁ = .9 h⁻¹

`e^(\lambda_2 t)` = 5
`e^(\lambda_1 t)`

Putting t = 2 h , λ₁ = .9 h⁻¹

`e^(\lambda_2* 2)` = 5
`e^(.9* 2)`

`e^(\lambda_2* 2)` = 30.25

2 x λ₂ = 3.41

λ₂ = 1.7047

Half life of B = .693 / 1.7047

= .4065 hours .

= .41 hours .