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:sqrt2`
D. `sqrt2:1`

Explanation

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

### Step-by-Step Solution:

1. **Understanding Kinetic Energy**:
The kinetic energy (KE) of a particle is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
For the two particles, we have:
- For mass `m`:
\[ KE_1 = \frac{1}{2} m v_1^2 \]
- For mass `2m`:
\[ KE_2 = \frac{1}{2} (2m) v_2^2 = m v_2^2 \]
Since both particles have equal kinetic energies:
\[ \frac{1}{2} m v_1^2 = m v_2^2 \]

2. **Simplifying the Equation**:
We can cancel `m` from both sides (assuming `m ≠ 0`):
\[ \frac{1}{2} v_1^2 = v_2^2 \]
Rearranging gives:
\[ v_1^2 = 2 v_2^2 \]
Taking the square root:
\[ v_1 = \sqrt{2} v_2 \]

3. **Finding the Momentum**:
The momentum (p) of a particle is given by:
\[ p = mv \]
Therefore, the momenta of the two particles are:
- For mass `m`:
\[ p_1 = m v_1 = m (\sqrt{2} v_2) = \sqrt{2} m v_2 \]
- For mass `2m`:
\[ p_2 = 2m v_2 \]

4. **Calculating the de Broglie Wavelength**:
The de Broglie wavelength (λ) is given by:
\[ \lambda = \frac{h}{p} \]
Thus, for the two particles:
- For mass `m`:
\[ \lambda_1 = \frac{h}{p_1} = \frac{h}{\sqrt{2} m v_2} \]
- For mass `2m`:
\[ \lambda_2 = \frac{h}{p_2} = \frac{h}{2m v_2} \]

5. **Finding the Ratio of Wavelengths**:
Now we can find the ratio of the de Broglie wavelengths:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{\sqrt{2} m v_2}}{\frac{h}{2m v_2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Final Answer:
The ratio of the de Broglie wavelengths of the two particles is:
\[ \frac{\lambda_1}{\lambda_2} = \sqrt{2} \]