A radioavtive source in the form of a metal sphere of daimeter `10^(-3)` m emits `beta`-particles at a constant rate of `6.25 xx 10^(10)` particles per second. If the source is electrically insulated, how long will it take for its potential to rise by `1.0 V`, assuming that `80%` of the emitted `beta`-particles escape the socurce?

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the radius of the sphere
Given the diameter of the sphere is \(10^{-3}\) m, the radius \(R\) can be calculated as:
\[ R = \frac{\text{Diameter}}{2} = \frac{10^{-3}}{2} = 0.5 \times 10^{-3} \text{ m} = 5 \times 10^{-4} \text{ m} \]

**Hint:** Remember that the radius is half of the diameter.

### Step 2: Calculate the charge needed to raise the potential by 1 V
Using the formula for electric potential \(V\):
\[ V = \frac{kQ}{R} \]
where \(k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\). Rearranging for \(Q\):
\[ Q = \frac{VR}{k} \]
Substituting the values:
\[ Q = \frac{(1 \, \text{V})(5 \times 10^{-4} \, \text{m})}{9 \times 10^9 \, \text{N m}^2/\text{C}^2} = \frac{5 \times 10^{-4}}{9 \times 10^9} = \frac{5}{9} \times 10^{-13} \, \text{C} \approx 0.555 \times 10^{-13} \, \text{C} \]

**Hint:** Use the formula for potential and rearrange it to find the charge \(Q\).

### Step 3: Calculate the number of beta particles needed to accumulate this charge
The charge of one beta particle (electron) is \(e = 1.6 \times 10^{-19} \, \text{C}\). The number of beta particles \(n\) needed can be calculated using:
\[ Q = n \cdot e \implies n = \frac{Q}{e} \]
Substituting the values:
\[ n = \frac{0.555 \times 10^{-13}}{1.6 \times 10^{-19}} \approx 3.47 \times 10^{6} \]

**Hint:** Remember that the charge is quantized, so you can find the number of particles by dividing the total charge by the charge of one particle.

### Step 4: Adjust for the percentage of beta particles that escape
Since 80% of the emitted beta particles escape, we need to find the effective number of beta particles emitted:
\[ \text{Effective } n = \frac{n}{0.8} = \frac{3.47 \times 10^{6}}{0.8} \approx 4.34 \times 10^{6} \]

**Hint:** Adjust the number of particles based on the percentage that escapes.

### Step 5: Calculate the time required for this number of beta particles to be emitted
The emission rate of beta particles is given as \(6.25 \times 10^{10} \, \text{particles/s}\). The time \(t\) required can be calculated using:
\[ t = \frac{\text{Total number of particles}}{\text{Rate}} = \frac{4.34 \times 10^{6}}{6.25 \times 10^{10}} \]
Calculating this gives:
\[ t \approx 6.94 \times 10^{-5} \, \text{s} = 69.4 \, \mu s \]

**Hint:** Use the formula for time as the total number of particles divided by the rate of emission.

### Final Answer
The time it will take for the potential to rise by 1.0 V is approximately \(69.4 \, \mu s\).