A radioactive substance .A. is being generated at a constant rate `C(100xx10^(6)` atoms /sec) it disintegrates at a rate of `lamda(37dps)` to form B. Initially there are no A and B atoms. If the number of atoms of B after one mean life of A is `1xx10^(x)` atoms, then find the value of x

Answer

To solve the problem step by step, we will analyze the generation and disintegration of the radioactive substance A and its conversion into substance B.

### Step 1: Define the constants
Let:
- \( C = 100 \times 10^6 \) atoms/second (rate of generation of A)
- \( \lambda = 37 \) disintegrations/second (disintegration constant of A)

### Step 2: Write the differential equation for A
The rate of change of the number of atoms of A can be expressed as:
\[ \frac{dA}{dt} = C - \lambda A \]
This equation states that the change in the number of atoms of A over time is equal to the rate of generation minus the rate of disintegration.

### Step 3: Solve the differential equation for A
Rearranging the equation gives:
\[ \frac{dA}{C - \lambda A} = dt \]
Integrating both sides, we get:
\[ \int \frac{dA}{C - \lambda A} = \int dt \]
This leads to:
\[ -\frac{1}{\lambda} \ln |C - \lambda A| = t + C_1 \]
where \( C_1 \) is the integration constant.

### Step 4: Apply initial conditions
Initially, at \( t = 0 \), \( A = 0 \):
\[ -\frac{1}{\lambda} \ln |C| = 0 + C_1 \implies C_1 = -\frac{1}{\lambda} \ln |C| \]
Substituting back, we have:
\[ -\frac{1}{\lambda} \ln |C - \lambda A| = t - \frac{1}{\lambda} \ln |C| \]
This simplifies to:
\[ C - \lambda A = C e^{-\lambda t} \]
Thus, we can express A as:
\[ A = \frac{C}{\lambda} (1 - e^{-\lambda t}) \]

### Step 5: Calculate A after one mean life
The mean life \( \tau \) of A is given by \( \tau = \frac{1}{\lambda} \).
After one mean life (\( t = \tau \)):
\[ A = \frac{C}{\lambda} (1 - e^{-1}) \approx \frac{C}{\lambda} (1 - 0.3679) \approx \frac{C}{\lambda} \times 0.6321 \]
Substituting \( C \) and \( \lambda \):
\[ A \approx \frac{100 \times 10^6}{37} \times 0.6321 \approx 1.71 \times 10^6 \text{ atoms} \]

### Step 6: Calculate the rate of formation of B
The rate of formation of B is given by:
\[ \frac{dB}{dt} = \lambda A \]
Integrating this from \( t = 0 \) to \( t = \tau \):
\[ B = \int_0^{\tau} \lambda A \, dt = \lambda \int_0^{\tau} A \, dt \]
Substituting \( A \):
\[ B = \lambda \int_0^{\tau} \frac{C}{\lambda} (1 - e^{-\lambda t}) \, dt \]
This simplifies to:
\[ B = C \left( t - \frac{1}{\lambda} e^{-\lambda t} \right) \Big|_0^{\tau} = C \left( \frac{1}{\lambda} - \frac{1}{\lambda} e^{-1} \right) \]
Calculating \( B \):
\[ B = C \left( \frac{1}{\lambda} (1 - e^{-1}) \right) \approx 100 \times 10^6 \left( \frac{1}{37} \times 0.6321 \right) \approx 1.71 \times 10^6 \text{ atoms} \]

### Step 7: Final calculation
We know that after one mean life, the number of atoms of B is given as \( 1 \times 10^x \) atoms. From our calculation:
\[ 1.71 \times 10^6 \approx 1 \times 10^6 \]
Thus, comparing this with \( 1 \times 10^x \):
\[ x = 6 \]

### Conclusion
The value of \( x \) is \( 6 \).