Which of these particles has the minimum De-Broglie wavelength, when the kinetic energy of all the particle is same?

Question

Accepted Answer

To determine which particle has the minimum De-Broglie wavelength when all particles have the same kinetic energy, we need to consider the De-Broglie wavelength formula:

$$
\lambda = \frac{h}{p}
$$

where $ \lambda $ is the De-Broglie wavelength, $ h $ is Planck's constant, and $ p $ is the momentum of the particle. The momentum $ p $ can be expressed in terms of kinetic energy $ KE $ as:

$$
p = \sqrt{2m \cdot KE}
$$

where $ m $ is the mass of the particle. Substituting this into the De-Broglie wavelength formula gives:

$$
\lambda = \frac{h}{\sqrt{2m \cdot KE}}
$$

From this equation, it is clear that the De-Broglie wavelength is inversely proportional to the square root of the mass of the particle when the kinetic energy is constant. Therefore, the particle with the greatest mass will have the smallest De-Broglie wavelength.

1. **Option 4: Alpha Particle** - An alpha particle consists of 2 protons and 2 neutrons, making it significantly more massive than the other particles listed. Given the same kinetic energy, the alpha particle's large mass results in the smallest De-Broglie wavelength. This is why Option 4 is correct.

2. **Option 1: Electron** - Electrons have a very small mass compared to alpha particles. Therefore, for the same kinetic energy, electrons will have a much larger De-Broglie wavelength than alpha particles.

3. **Option 2: Photon** - Photons are massless particles. The concept of De-Broglie wavelength for photons is different because they always travel at the speed of light and their wavelength is determined by their energy, not kinetic energy in the classical sense. Thus, they do not fit the same analysis as massive particles.

4. **Option 3: Neutron** - Neutrons are more massive than electrons but less massive than alpha particles. Therefore, while they have a smaller De-Broglie wavelength than electrons for the same kinetic energy, their wavelength is still larger than that of alpha particles.

In conclusion, the alpha particle, being the most massive among the given options, has the minimum De-Broglie wavelength when all particles have the same kinetic energy. This supports Option 4 as the correct answer.