If alpha particle, proton and electron move with the same momentum, them their respective de Broglie wavelengths `lamda_(alpha),lamda_(p),lamda_(e)` are related as

Options

A. `lamda_(alpha)=lamda_(p)=lamda_(e)`
B. `lamda_(alpha)ltlamda_(p)ltlamda_(e)`
C. `lamda_(alpha)gtlamda_(p)gtlamda_(e)`
D. `lamda_(p)gtlamda_(e)gtlamda_(alpha)`

Explanation

To solve the problem, we need to establish the relationship between the de Broglie wavelengths of an alpha particle, a proton, and an electron when they all have the same momentum.

### Step-by-Step Solution:

1. **Understand the de Broglie Wavelength Formula**:
The de Broglie wavelength (λ) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is the Planck constant and \( p \) is the momentum of the particle.

2. **Identify the Given Condition**:
We know that the momentum of the alpha particle, proton, and electron is the same. Let's denote this common momentum as \( p \).

3. **Write the Wavelengths for Each Particle**:
Using the de Broglie wavelength formula, we can express the wavelengths for each particle:
- For the alpha particle:
\[ \lambda_{\alpha} = \frac{h}{p} \]
- For the proton:
\[ \lambda_{p} = \frac{h}{p} \]
- For the electron:
\[ \lambda_{e} = \frac{h}{p} \]

4. **Compare the Wavelengths**:
Since all three particles have the same momentum \( p \), we can see that:
\[ \lambda_{\alpha} = \lambda_{p} = \lambda_{e} \]

5. **Conclusion**:
Therefore, the relationship between their respective de Broglie wavelengths is:
\[ \lambda_{\alpha} : \lambda_{p} : \lambda_{e} = 1 : 1 : 1 \]

### Final Answer:
The de Broglie wavelengths of the alpha particle, proton, and electron are equal when they have the same momentum.

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