Consider a point sorce emiiting `alpha`- particles and receptor and receptor of area `1 cm^(2)` placed `1m` away from source. Receptor records any `alpha`-particle falling on it. If the contains `N_(0) = 3.0 xx 10^(16)` active nuclie and the receptor records a rate of `A = 50000 "counts"//"second"`. Assume that the source emits alpha particles unifomrly in all direction and the alpha particles fall nearly normally on the window. If decay constant is `3n xx 10^(-(n+1))`, then the value of `n`

Explanation

To solve the problem, we need to find the value of \( n \) given the decay constant \( \lambda = 3 \times 10^{-(n+1)} \) and the information about the alpha particles emitted by a point source.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Number of active nuclei, \( N_0 = 3.0 \times 10^{16} \)
- Rate of counts recorded by the receptor, \( A = 50000 \) counts/second
- Distance from the source to the receptor, \( r = 1 \) m
- Area of the receptor, \( A_r = 1 \) cm² = \( 1 \times 10^{-4} \) m²

2. **Calculate Total Surface Area of Emission:**
The total surface area \( S \) from which the alpha particles are emitted is given by the formula for the surface area of a sphere:
\[ S = 4\pi r^2 \]
Substituting \( r = 1 \) m:
\[ S = 4\pi (1)^2 = 4\pi \approx 12.56 \, \text{m}^2 \]

3. **Calculate the Total Number of Counts Emitted:**
The number of counts emitted per second by the source can be calculated using the ratio of the area of the receptor to the total surface area:
\[ \text{Total counts emitted per second} = A \times \frac{S}{A_r} \]
Substituting the values:
\[ \text{Total counts emitted per second} = 50000 \times \frac{12.56}{1 \times 10^{-4}} = 50000 \times 125600 = 6.28 \times 10^9 \, \text{counts/second} \]

4. **Relate Counts to Decay Constant:**
We know that the decay rate can also be expressed as:
\[ \frac{dN}{dt} = \lambda N_0 \]
Setting this equal to the total counts emitted:
\[ \lambda N_0 = 6.28 \times 10^9 \]
Substituting \( N_0 = 3.0 \times 10^{16} \):
\[ \lambda \times 3.0 \times 10^{16} = 6.28 \times 10^9 \]
Solving for \( \lambda \):
\[ \lambda = \frac{6.28 \times 10^9}{3.0 \times 10^{16}} = 2.0933 \times 10^{-7} \, \text{s}^{-1} \]

5. **Relate Decay Constant to \( n \):**
We know from the problem that:
\[ \lambda = 3 \times 10^{-(n+1)} \]
Setting the two expressions for \( \lambda \) equal:
\[ 2.0933 \times 10^{-7} = 3 \times 10^{-(n+1)} \]
Rearranging gives:
\[ 10^{-(n+1)} = \frac{2.0933 \times 10^{-7}}{3} \approx 6.9777 \times 10^{-8} \]
Taking logarithm base 10:
\[ -(n+1) = \log_{10}(6.9777 \times 10^{-8}) \]
Calculating the logarithm:
\[ -(n+1) \approx -7.155 \]
Thus:
\[ n + 1 \approx 7.155 \implies n \approx 6.155 \]

6. **Final Calculation for \( n \):**
Rounding to the nearest integer gives \( n = 7 \).

### Final Answer:
The value of \( n \) is \( 7 \).

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