Question

The following nonrelativistic particles all have the same kinetic energy. Rank them in order of their de Broglie wavelengths, greatest first: electron, alpha particle, neutron.

Short answer

The rank is electron, neutron, alpha particle.

Step-by-step solution

Describe the de Broglie wavelength

The de Broglie wavelength is given by,

$\mathit{\lambda }\mathbf{=}\frac{\mathbf{h}}{\mathbf{m}\mathbf{v}}$

Here, h is the plank’s constant, m is the mass and v is the velocity.

Rank the de Broglie wavelengths for electron, alpha particle, and neutron

From the de Broglie wavelength, it can be observed that the mass of the particle is inversely proportional to the wavelength.

It is known that the mass of the neutron is $m_n=1.67\times {10}^{-27}\text{kg}$, the mass of alpha particle is $m_{\alpha }=6.664\times {10}^{-27}\text{kg}$ and the mass of the electron is $m_e=9.11\times {10}^{-31}\text{kg}$.Clearly, $m_e<m_n<m_{\alpha }$. So, rank of de Broglie wavelength is ${\lambda }_{\text{e}}>{\lambda }_n>{\lambda }_{\alpha }$.

Therefore, the rank is electron, neutron, alpha particle.