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

Answer:

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

$e^{\lambda_2\times 2}$ = 5  $e^{.9\times 2}$

$e^{\lambda_2\times 2}$ = 30.25

2 x λ₂ = 3.41

λ₂ = 1.7047

Half life of B = .693 / 1.7047

= .4065 hours .

= .41 hours .