A charged particle moving in a uniform magnetic field penetrates a layer of lead and there by loses one-half of its kinetic energy. How does the radius of curvature of its path change?

Options

A. (a)The radius reduces to `r sqrt(2)`
B. (b)The radius reduces to `r/sqrt2`
C. (c)The radius remains the same
D. (d)The radius becomes `r//2`

Answer

To solve the problem, we need to analyze how the radius of curvature of a charged particle's path in a magnetic field changes when it loses half of its kinetic energy after penetrating a layer of lead.

### Step-by-Step Solution:

1. **Understanding the Kinetic Energy**:
The kinetic energy (K.E) of a charged particle is given by the formula:
\[ K.E = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the particle and \( v \) is its velocity.

2. **Initial Kinetic Energy**:
Let the initial kinetic energy be \( K \). Thus, we have:
\[ K = \frac{1}{2} mv^2 \]

3. **Final Kinetic Energy**:
After penetrating the lead, the particle loses half of its kinetic energy, so the final kinetic energy \( K' \) is:
\[ K' = \frac{1}{2} K = \frac{1}{4} mv^2 \]

4. **Finding the Final Velocity**:
From the final kinetic energy, we can express the final velocity \( v' \):
\[ K' = \frac{1}{2} mv'^2 \implies \frac{1}{4} mv^2 = \frac{1}{2} mv'^2 \]
Simplifying this gives:
\[ v'^2 = \frac{1}{2} v^2 \implies v' = \frac{v}{\sqrt{2}} \]

5. **Radius of Curvature in a Magnetic Field**:
The radius of curvature \( r \) of a charged particle moving in a magnetic field \( B \) is given by:
\[ r = \frac{mv}{qB} \]
where \( q \) is the charge of the particle.

6. **Initial Radius of Curvature**:
The initial radius of curvature \( r \) can be expressed as:
\[ r = \frac{mv}{qB} \]

7. **Final Radius of Curvature**:
The final radius of curvature \( r' \) after the particle's velocity changes to \( v' \) is:
\[ r' = \frac{mv'}{qB} \]
Substituting \( v' = \frac{v}{\sqrt{2}} \):
\[ r' = \frac{m \left(\frac{v}{\sqrt{2}}\right)}{qB} = \frac{mv}{\sqrt{2} qB} = \frac{r}{\sqrt{2}} \]

8. **Conclusion**:
Therefore, the radius of curvature of the particle's path decreases to:
\[ r' = \frac{r}{\sqrt{2}} \]

### Final Answer:
The radius of curvature decreases to \( \frac{r}{\sqrt{2}} \).