Two particles of masses `m` and `2m` have equal kinetic energies. Their de Broglie wavelengths area in the ratio of:

Options

A. `1:1`
B. `1:2`
C. `1:sqrt(2)`
D. `sqrt(2):1`

Answer

To solve the problem, we need to find the ratio of the de Broglie wavelengths of two particles with masses `m` and `2m` that have equal kinetic energies.

### Step-by-step Solution:

1. **Understand the de Broglie wavelength formula**:
The de Broglie wavelength (λ) of a particle is given by the formula:
\[ \lambda = \frac{h}{\sqrt{2mk}} \]
where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the particle,
- \( k \) is the kinetic energy of the particle.

2. **Set up the problem**:
Let’s denote the two particles:
- Particle 1 with mass \( m \) and kinetic energy \( k \)
- Particle 2 with mass \( 2m \) and the same kinetic energy \( k \)

3. **Write the de Broglie wavelength for both particles**:
For Particle 1:
\[ \lambda_1 = \frac{h}{\sqrt{2 \cdot m \cdot k}} \]
For Particle 2:
\[ \lambda_2 = \frac{h}{\sqrt{2 \cdot (2m) \cdot k}} = \frac{h}{\sqrt{4mk}} \]

4. **Simplify the expression for λ2**:
We can rewrite λ2 as:
\[ \lambda_2 = \frac{h}{2\sqrt{mk}} \]

5. **Find the ratio of the de Broglie wavelengths**:
Now, we can find the ratio \( \frac{\lambda_1}{\lambda_2} \):
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{\sqrt{2mk}}}{\frac{h}{2\sqrt{mk}}} \]

The \( h \) cancels out:
\[ \frac{\lambda_1}{\lambda_2} = \frac{2\sqrt{mk}}{\sqrt{2mk}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

6. **Final result**:
Thus, the ratio of the de Broglie wavelengths of the two particles is:
\[ \frac{\lambda_1}{\lambda_2} = \sqrt{2} : 1 \]

### Conclusion:
The ratio of the de Broglie wavelengths of the two particles is \( \sqrt{2} : 1 \).