A radio nuclide with half life `T = 69.31` second emits `beta`-particles of average kinetic energy `E = 11.25eV`. At an instant concentration of `beta`-particles at distance, `r = 2m` from nuclide is `n = 3 xx 10^(13)` per `m^(3)`.
(i) Calculate number of nuclei in the nuclide at that instant.
(ii) If a small circular plate is placed at distance `r` from nuclide such that `beta`-particles strike the plate normally and come to rest, calculate pressure experienced by the plate due to collision of `beta`-particle. (Mass of `beta`-particle `= 9 xx 10^(-31)kg`) `(log_(e) 2 = 0.693)`

Explanation

To solve the problem step by step, we will break it down into two parts as specified in the question.

### Part (i): Calculate the number of nuclei in the nuclide at that instant.

1. **Calculate the decay constant (λ)**:
The decay constant can be calculated using the half-life formula:
\[ \lambda = \frac{0.693}{T_{1/2}} \]
Given \( T_{1/2} = 69.31 \, \text{s} \),
\[ \lambda = \frac{0.693}{69.31} \approx 0.01 \, \text{s}^{-1} \]

2. **Use the relationship between activity (A), decay constant (λ), and number of nuclei (N)**:
The activity of the nuclide is given by:
\[ A = \lambda N \]

3. **Relate the concentration of beta particles (n) to the activity (A)**:
The concentration of beta particles at distance \( r \) is given as:
\[ n = 3 \times 10^{13} \, \text{m}^{-3} \]
The number of beta particles emitted per second can be expressed as:
\[ A = n \cdot \text{Area} \cdot \text{Velocity} \]
The area for a sphere of radius \( r \) is \( 4\pi r^2 \) and the thickness is negligible, thus:
\[ A = n \cdot 4\pi r^2 \cdot v \]

4. **Calculate the velocity (v) of beta particles**:
The kinetic energy (E) of beta particles is given as \( E = 11.25 \, \text{eV} \). Convert this to Joules:
\[ E = 11.25 \times 1.6 \times 10^{-19} \, \text{J} = 1.8 \times 10^{-18} \, \text{J} \]
Using the kinetic energy formula:
\[ E = \frac{1}{2} mv^2 \implies v = \sqrt{\frac{2E}{m}} \]
Substituting \( m = 9 \times 10^{-31} \, \text{kg} \):
\[ v = \sqrt{\frac{2 \times 1.8 \times 10^{-18}}{9 \times 10^{-31}}} \approx 1.83 \times 10^{6} \, \text{m/s} \]

5. **Substitute values into the activity equation**:
Now substituting \( n \), \( r \), and \( v \) into the activity equation:
\[ A = n \cdot 4\pi (2)^2 \cdot v \]
\[ A = 3 \times 10^{13} \cdot 4\pi \cdot 4 \cdot 1.83 \times 10^{6} \]
Calculate \( A \):
\[ A \approx 3 \times 10^{13} \cdot 50.27 \cdot 1.83 \times 10^{6} \approx 2.77 \times 10^{21} \, \text{decays/s} \]

6. **Calculate the number of nuclei (N)**:
Using \( A = \lambda N \):
\[ N = \frac{A}{\lambda} = \frac{2.77 \times 10^{21}}{0.01} \approx 2.77 \times 10^{23} \]

### Part (ii): Calculate pressure experienced by the plate due to collision of beta particles.

1. **Calculate the number of beta particles striking the plate**:
The number of beta particles striking the plate per second is given by:
\[ N_{\text{strikes}} = \frac{A \cdot S}{4\pi r^2} \]
Where \( S \) is the area of the plate.

2. **Calculate the momentum change per beta particle**:
The change in momentum for one beta particle is:
\[ \Delta p = mv \]

3. **Calculate the total force on the plate**:
The total force due to the beta particles striking the plate is:
\[ F = N_{\text{strikes}} \cdot \Delta p \]

4. **Calculate the pressure (P)**:
Pressure is defined as force per unit area:
\[ P = \frac{F}{S} \]

5. **Substituting values**:
Substitute the expressions into the pressure formula:
\[ P = \frac{N_{\text{strikes}} \cdot mv}{S} \]

6. **Final calculation**:
After substituting all known values, we find:
\[ P \approx 1.08 \times 10^{-4} \, \text{N/m}^2 \]