To solve the problem, we need to find the ratio of the radii of the circular paths of a proton, a deuteron, and an alpha particle when they enter a uniform magnetic field with the same kinetic energy.
### Step-by-Step Solution:
1. **Understanding the particles**:
- **Proton (p)**: Mass = \( m_p = 1u \), Charge = \( q_p = +e \)
- **Deuteron (d)**: Mass = \( m_d = 2u \), Charge = \( q_d = +e \)
- **Alpha particle (α)**: Mass = \( m_\alpha = 4u \), Charge = \( q_\alpha = +2e \)
2. **Kinetic Energy**:
Since all particles have the same kinetic energy (KE), we can express the kinetic energy as:
\[
KE = \frac{1}{2} mv^2
\]
For each particle, we can set this equal to a constant \( K \):
\[
K = \frac{1}{2} m_p v_p^2 = \frac{1}{2} m_d v_d^2 = \frac{1}{2} m_\alpha v_\alpha^2
\]
3. **Finding the radius of circular motion**:
When a charged particle moves in a magnetic field, the radius \( R \) of its circular path is given by:
\[
R = \frac{mv}{qB}
\]
Here, \( m \) is the mass, \( v \) is the velocity, \( q \) is the charge, and \( B \) is the magnetic field strength.
4. **Expressing velocity in terms of kinetic energy**:
From the kinetic energy equation, we can express velocity \( v \) as:
\[
v = \sqrt{\frac{2K}{m}}
\]
5. **Substituting for radius**:
Substituting \( v \) into the radius formula:
\[
R = \frac{m \sqrt{\frac{2K}{m}}}{qB} = \frac{\sqrt{2Km}}{qB}
\]
6. **Finding the ratio of radii**:
Now we can find the ratio of the radii for the three particles:
\[
R_p = \frac{\sqrt{2K m_p}}{q_p B}, \quad R_d = \frac{\sqrt{2K m_d}}{q_d B}, \quad R_\alpha = \frac{\sqrt{2K m_\alpha}}{q_\alpha B}
\]
The ratios of the radii will be:
\[
\frac{R_p}{R_d} = \frac{\sqrt{m_p}/q_p}{\sqrt{m_d}/q_d}, \quad \frac{R_d}{R_\alpha} = \frac{\sqrt{m_d}/q_d}{\sqrt{m_\alpha}/q_\alpha}
\]
Therefore, we can express the ratio \( R_p : R_d : R_\alpha \) as:
\[
R_p : R_d : R_\alpha = \frac{\sqrt{m_p}}{q_p} : \frac{\sqrt{m_d}}{q_d} : \frac{\sqrt{m_\alpha}}{q_\alpha}
\]
7. **Substituting values**:
- For the proton: \( m_p = 1u, q_p = e \)
- For the deuteron: \( m_d = 2u, q_d = e \)
- For the alpha particle: \( m_\alpha = 4u, q_\alpha = 2e \)
Thus,
\[
R_p : R_d : R_\alpha = \frac{\sqrt{1u}}{e} : \frac{\sqrt{2u}}{e} : \frac{\sqrt{4u}}{2e}
\]
Simplifying this gives:
\[
R_p : R_d : R_\alpha = 1 : \sqrt{2} : 1
\]
8. **Final Ratio**:
Therefore, the final ratio of the radii of their circular paths is:
\[
R_p : R_d : R_\alpha = 1 : \sqrt{2} : 1
\]
