Deutron and alpha particle having same K.E. in magnetic field. If the ratio of radius of Deutron and alpha particle is `xsqrt2`. Then x=?

Answer

To solve the problem, we need to find the value of \( x \) given that the ratio of the radii of a deuteron and an alpha particle in a magnetic field is \( x \sqrt{2} \) and that both particles have the same kinetic energy.

### Step-by-step Solution:

1. **Understanding the Radius Formula**:
The radius \( r \) of a charged particle moving in a magnetic field is given by the formula:
\[ r = \frac{mv}{qB} \]
where \( m \) is the mass of the particle, \( v \) is its velocity, \( q \) is the charge, and \( B \) is the magnetic field strength.

2. **Setting Up the Ratio**:
For the deuteron and alpha particle, we can write:
\[ \frac{R_{deuteron}}{R_{alpha}} = \frac{\frac{m_{deuteron} v_{deuteron}}{q_{deuteron} B}}{\frac{m_{alpha} v_{alpha}}{q_{alpha} B}} = \frac{m_{deuteron} v_{deuteron} q_{alpha}}{m_{alpha} v_{alpha} q_{deuteron}} \]

3. **Using Kinetic Energy**:
Since both particles have the same kinetic energy \( K \), we can express the velocity in terms of kinetic energy:
\[ K = \frac{1}{2} mv^2 \implies v = \sqrt{\frac{2K}{m}} \]
Therefore, for the deuteron and alpha particle:
\[ v_{deuteron} = \sqrt{\frac{2K}{m_{deuteron}}}, \quad v_{alpha} = \sqrt{\frac{2K}{m_{alpha}}} \]

4. **Substituting Velocities**:
Substituting these velocities back into the radius ratio:
\[ \frac{R_{deuteron}}{R_{alpha}} = \frac{m_{deuteron} \sqrt{\frac{2K}{m_{deuteron}}} q_{alpha}}{m_{alpha} \sqrt{\frac{2K}{m_{alpha}}} q_{deuteron}} \]
The \( \sqrt{2K} \) terms cancel out:
\[ = \frac{m_{deuteron} \sqrt{m_{alpha}} q_{alpha}}{m_{alpha} \sqrt{m_{deuteron}} q_{deuteron}} \]

5. **Substituting Masses and Charges**:
The mass of the deuteron \( m_{deuteron} = 2 \) (in atomic mass units), and the mass of the alpha particle \( m_{alpha} = 4 \). The charge of the deuteron \( q_{deuteron} = 1 \) (in elementary charge units), and the charge of the alpha particle \( q_{alpha} = 2 \).
Substituting these values:
\[ \frac{R_{deuteron}}{R_{alpha}} = \frac{2 \sqrt{4} \cdot 2}{4 \cdot \sqrt{2} \cdot 1} = \frac{4}{2\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

6. **Equating to Given Ratio**:
We are given that:
\[ \frac{R_{deuteron}}{R_{alpha}} = x \sqrt{2} \]
Thus, we have:
\[ \sqrt{2} = x \sqrt{2} \]

7. **Solving for \( x \)**:
Dividing both sides by \( \sqrt{2} \):
\[ x = 1 \]

### Final Answer:
\[ x = 1 \]