Five particles, proton, neutron, `beta`-particle, electron and `alpha`-particle are moving with speed v.

Options

A. de Broglie wavelength of `beta`-particle and `alpha`-particle is longest
B. de Broglie wavelength of `beta`-proton and neutron is longest
C. de Broglie wavelength of `beta`-particle and electron is longest
D. de Broglie wavelength of electron and neutron is longest

Explanation

To solve the problem of comparing the de Broglie wavelengths of five particles (proton, neutron, beta particle, electron, and alpha particle) moving with the same speed \( v \), we will follow these steps:

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

### Step 2: Express momentum in terms of mass and velocity
The momentum \( p \) of a particle can be expressed as:
\[ p = m \cdot v \]
where \( m \) is the mass of the particle and \( v \) is its velocity. Substituting this into the de Broglie wavelength formula gives:
\[ \lambda = \frac{h}{m \cdot v} \]

### Step 3: Analyze the given conditions
Since all five particles are moving with the same speed \( v \), the only variable affecting the de Broglie wavelength will be the mass \( m \) of each particle. Therefore, we can rewrite the wavelength as:
\[ \lambda \propto \frac{1}{m} \]
This means that the de Broglie wavelength is inversely proportional to the mass of the particle.

### Step 4: List the masses of the particles
We need to know the masses of the particles involved:
- Mass of proton \( m_p = 1 \, \text{amu} \)
- Mass of neutron \( m_n = 1 \, \text{amu} \)
- Mass of beta particle \( m_{\beta} \approx 0 \, \text{amu} \) (similar to electron)
- Mass of electron \( m_e \approx 0 \, \text{amu} \)
- Mass of alpha particle \( m_{\alpha} = 4 \, \text{amu} \)

### Step 5: Compare the masses and deduce the wavelengths
From the masses, we can determine the order of the de Broglie wavelengths:
- Beta particle and electron have the smallest mass (approximately 0 amu), so they will have the largest de Broglie wavelengths.
- Proton and neutron have the same mass (1 amu), so their de Broglie wavelengths will be smaller than those of the beta particle and electron but larger than that of the alpha particle.
- The alpha particle has the largest mass (4 amu), so it will have the smallest de Broglie wavelength.

### Step 6: Summarize the order of de Broglie wavelengths
Based on the analysis:
1. \( \lambda_{\beta} \) and \( \lambda_{e} \) (largest)
2. \( \lambda_{p} \) and \( \lambda_{n} \)
3. \( \lambda_{\alpha} \) (smallest)

### Step 7: Conclusion
The correct order of the de Broglie wavelengths from longest to shortest is:
\[ \lambda_{\beta} = \lambda_{e} > \lambda_{p} = \lambda_{n} > \lambda_{\alpha} \]