A proton, a neutron, an electron and an $$\alpha$$-particle have same energy. If $$\lambda_p, \lambda_n, \lambda_e$$ and $$\lambda_\alpha$$ are the de Broglie's wavelengths of proton, neutron, electron and $$\alpha$$ particle respectively, then choose the correct relation from the following

The de Broglie wavelength is given by:

$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$$

where $$h$$ is Planck's constant, $$m$$ is the mass of the particle, and $$E$$ is the kinetic energy.

Since all particles have the same energy $$E$$:

$$\lambda \propto \frac{1}{\sqrt{m}}$$

The larger the mass, the smaller the wavelength.

Comparing masses:

- Electron: $$m_e$$ (smallest mass)

- Proton: $$m_p \approx 1836\, m_e$$

- Neutron: $$m_n \approx m_p$$ (slightly greater than proton mass)

- Alpha particle: $$m_\alpha \approx 4m_p$$ (largest mass)

Since $$m_e < m_p < m_n < m_\alpha$$:

$$\lambda_e > \lambda_p > \lambda_n > \lambda_\alpha$$

This can be rewritten as:

$$\lambda_\alpha < \lambda_n < \lambda_p < \lambda_e$$

Hence, the correct answer is Option B.