Question

In a radioactive decay process, the activity is defined as A = - \(\frac {dN}{dt}\), where 𝑁(𝑡) is the number of radioactive nuclei at time 𝑡. Two radioactive sources, 𝑆₁ and 𝑆₂ have same activity at time 𝑡 = 0. At a later time, the activities of 𝑆₁ and 𝑆₂ are 𝐴₁ and 𝐴₂, respectively. When 𝑆₁ and 𝑆₂ have just completed their 3^rd and 7^th half-lives, respectively, the ratio \(\frac{A_1}{A_2}\) is _________.

Answer 1

The number of radioactive nuclei N(t) at a given time t is given by:
N(t) = N₀ \(\times\)exp(-λ_t)
The activity A of a radioactive source is defined as the rate of decay, which is the derivative of the number of radioactive nuclei with respect to time: A = -\(\frac {dN}{dt}\) = λ\(\times\) N(t) 
Given that the activities of sources S₁ and S₂ are A₁ and A₂, respectively, we can write: 
A₁ = λ₁ \(\times\) N₁(t) 
A₂ = λ₂ \(\times\) N₂(t) 
After the completion of the 3^rd half-life for source S₁, the remaining number of radioactive nuclei is: 
N₁(t) = N₀ \(\times\) (\(\frac{1}{2}\))³ = N₀ \(\times\) \(\frac {1}{8}\) 
Similarly, after the completion of the 7^th half-life for source S₂, the remaining number of radioactive nuclei is: 
N₂(t) = N₀ \(\times\) \([\frac{1}{2}]\)⁷ = N₀ \(\times\) \(\frac {1}{128}\)
Now, let's substitute these values into the equations for activities A₁ and A₂ : 
A₁ = λ₁ \(\times\) N₀ \(\times\) (\(\frac {1}{8}\)) 

A₂ = λ₂ \(\times\) N₀ \(\times\) (\(\frac {1}{128}\)) 

To find the ratio \(\frac{A_1}{A_2}\), we can simplify: 

(\(\frac{A_1}{A_2}\)) = (\(\frac{\lambda_1}{\lambda_2}\)) \(\times\) (\(\frac {1}{8}\) ) \(\times\) (\(\frac {128}{1}\)) 
Since both sources have the same activity at time t = 0, we can assume their decay constants are the same (λ₁ = λ₂). Therefore, the ratio simplifies to: 
(\(\frac{A_1}{A_2}\)) = \(\frac {1}{8}\) \(\times\) \(\frac {128}{1}\) = \(\frac {128}{8}\) = 16 

Therefore, the ratio \(\frac{A_1}{A_2}\) is 16.

So, the answer is \(16\).

Answer 2

Given
S₁​ has just completed its 3rd half-life.
S₂​ has just completed its 7th half-life.
Initially, both sources have the same activity, \(A(0) = \lambda N_0\)
Calculating Activities After Given Half-Lives:
For S₁​ after 3 half-lives:
\(N_1 = N_0 \left( \frac{1}{2} \right)^3 = N_0 \left( \frac{1}{8} \right) = \frac{N_0}{8}\)
Hence, the activity A1A_1A1​ at this time is:
\(A_1 = \lambda N_1 = \lambda \left( \frac{N_0}{8} \right) = \frac{\lambda N_0}{8}\)
For S₂​ after 7 half-lives:
\(N_2 = N_0 \left( \frac{1}{2} \right)^7 = N_0 \left( \frac{1}{128} \right) = \frac{N_0}{128}\)
Hence, the activity A₂​ at this time is:
\(A_2 = \lambda N_2 = \lambda \left( \frac{N_0}{128} \right) = \frac{\lambda N_0}{128}\)
Calculating the Ratio of Activities:
The ratio \(\frac{A_1}{A_2}\)​ is:
\(\frac{A_1}{A_2} = \frac{\frac{\lambda N_0}{8}}{\frac{\lambda N_0}{128}} = \frac{\frac{1}{8}}{\frac{1}{128}} = \frac{128}{8} = 16\)

So, the answer is 16.