A radioactive source, in the form of a metallic sphere of radius `10^(-2)m` emits `beta-`particles at the rate of `5xx10^(10)` particles per second. The source is electrically insulated. How long will it take for its potential to be raised by `2V`, assuming that `40%` of the emitted `beta-`particles escape the source.

Explanation

To solve the problem step by step, we will follow the logic presented in the video transcript.

### Step 1: Understand the problem
We have a metallic sphere of radius \( r = 10^{-2} \, \text{m} \) that emits \( \beta^- \) particles at a rate of \( R = 5 \times 10^{10} \) particles per second. We need to find out how long it will take for the potential of the sphere to increase by \( 2 \, \text{V} \), given that \( 40\% \) of the emitted particles escape.

### Step 2: Calculate the charge required to raise the potential by \( 2 \, \text{V} \)
The formula for the electric potential \( V \) of a charged sphere is given by:

\[ V = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r} \]

Where:
- \( V \) is the potential,
- \( Q \) is the charge,
- \( r \) is the radius of the sphere,
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

We want to find \( Q \) when \( V = 2 \, \text{V} \):

\[ 2 = \frac{1}{4 \pi (8.85 \times 10^{-12})} \frac{Q}{10^{-2}} \]

Rearranging the formula to find \( Q \):

\[ Q = 2 \cdot 4 \pi (8.85 \times 10^{-12}) \cdot (10^{-2}) \]

Calculating \( Q \):

\[ Q = 2 \cdot 4 \cdot 3.14 \cdot (8.85 \times 10^{-12}) \cdot (10^{-2}) \approx 2 \cdot 1.11 \times 10^{-12} \approx 2.22 \times 10^{-12} \, \text{C} \]

### Step 3: Calculate the number of emitted particles that contribute to the charge
Since only \( 40\% \) of the emitted \( \beta^- \) particles escape, \( 60\% \) remain and contribute to the charge. The number of particles emitted in time \( t \) is:

\[ n = R \cdot t = 5 \times 10^{10} \cdot t \]

The charge contributed by these particles is:

\[ Q_{\text{total}} = n \cdot e = (R \cdot t) \cdot e \]

Where \( e \) (the charge of one \( \beta^- \) particle) is approximately \( 1.6 \times 10^{-19} \, \text{C} \). Since only \( 60\% \) of the particles contribute to the charge, we have:

\[ Q_{\text{total}} = 0.6 \cdot (R \cdot t) \cdot e \]

Substituting \( R \):

\[ Q_{\text{total}} = 0.6 \cdot (5 \times 10^{10} \cdot t) \cdot (1.6 \times 10^{-19}) \]

### Step 4: Set up the equation
Now we set the charge required to raise the potential equal to the total charge contributed by the particles:

\[ 2.22 \times 10^{-12} = 0.6 \cdot (5 \times 10^{10} \cdot t) \cdot (1.6 \times 10^{-19}) \]

### Step 5: Solve for \( t \)
Rearranging the equation to solve for \( t \):

\[ t = \frac{2.22 \times 10^{-12}}{0.6 \cdot (5 \times 10^{10} \cdot 1.6 \times 10^{-19})} \]

Calculating the denominator:

\[ 0.6 \cdot (5 \times 10^{10} \cdot 1.6 \times 10^{-19}) = 0.6 \cdot 8 \times 10^{-9} = 4.8 \times 10^{-9} \]

Now substituting back into the equation for \( t \):

\[ t = \frac{2.22 \times 10^{-12}}{4.8 \times 10^{-9}} \approx 4.63 \times 10^{-4} \, \text{s} \]

### Final Answer
Thus, the time required for the potential to be raised by \( 2 \, \text{V} \) is approximately:

\[ t \approx 4.63 \times 10^{-4} \, \text{s} \]