Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Discover detailed solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
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 [tex]e^{-\lambda t}[/tex]
N is no of atoms after time t , λ is decay constant and t is time .
For A
N(A) = N(A)₀ [tex]e^{-\lambda_1 t}[/tex]
For B
N(B) = N(B)₀ [tex]e^{-\lambda_2 t}[/tex]
N(A) = N(B) , for t = 2 h
N(A)₀ [tex]e^{-\lambda_1 t}[/tex] = N(B)₀ [tex]e^{-\lambda_2 t}[/tex]
N(A)₀ [tex]e^{-\lambda_1 t}[/tex] = 5 x N₀(A) [tex]e^{-\lambda_2 t}[/tex]
[tex]e^{-\lambda_1 t}[/tex] = 5 [tex]e^{-\lambda_2 t}[/tex]
[tex]e^{\lambda_2 t}[/tex] = 5 [tex]e^{\lambda_1 t}[/tex]
half life = .693 / λ
For A
.77 = .693 / λ₁
λ₁ = .9 h⁻¹
[tex]e^{\lambda_2 t}[/tex] = 5 [tex]e^{\lambda_1 t}[/tex]
Putting t = 2 h , λ₁ = .9 h⁻¹
[tex]e^{\lambda_2\times 2}[/tex] = 5 [tex]e^{.9\times 2}[/tex]
[tex]e^{\lambda_2\times 2}[/tex] = 30.25
2 x λ₂ = 3.41
λ₂ = 1.7047
Half life of B = .693 / 1.7047
= .4065 hours .
= .41 hours .
We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
