If particles are moving with same velocity , then maximum de - Broglie wavelength will be for

Options

A. proton
B. `alpha` particle
C. neutron
D. `beta`particles

Answer

To find out which particle has the maximum de Broglie wavelength when particles are moving with the same velocity, we can follow these steps:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the particle,
- \( v \) is the velocity of the particle.

### Step 2: Analyze the relationship
From the formula, we can see that the de Broglie wavelength is inversely proportional to the mass of the particle when the velocity is constant:
\[ \lambda \propto \frac{1}{m} \]
This means that as the mass \( m \) decreases, the wavelength \( \lambda \) increases.

### Step 3: Compare the masses of different particles
We need to compare the masses of the particles involved:
- Mass of a proton (\( m_p \)) is approximately \( 1.67 \times 10^{-27} \) kg.
- Mass of a neutron (\( m_n \)) is approximately \( 1.68 \times 10^{-27} \) kg.
- Mass of an alpha particle (\( m_{\alpha} \)) is approximately \( 6.64 \times 10^{-27} \) kg (which is about four times the mass of a proton).
- Mass of a beta particle (which is an electron, \( m_e \)) is approximately \( 9.11 \times 10^{-31} \) kg.

### Step 4: Determine the particle with the minimum mass
Among these particles:
- The mass of the beta particle (electron) is the smallest.
- The mass of the proton and neutron are similar and greater than that of the beta particle.
- The mass of the alpha particle is the greatest.

### Step 5: Conclude the particle with maximum de Broglie wavelength
Since the de Broglie wavelength is inversely proportional to mass, the particle with the smallest mass will have the maximum de Broglie wavelength. Therefore, the beta particle (electron) will have the maximum de Broglie wavelength when all particles are moving with the same velocity.

### Final Answer:
The maximum de Broglie wavelength will be for the beta particle (electron).

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