Question

A proton and an electron have the same kinetic energy. Which one has greater de Broglie wavelength and why?

Hint:

For the same kinetic energy, the particle with the smaller mass has a greater de Broglie wavelength.

Answer 1

The de Broglie wavelength (\(\lambda\)) of a particle is given by the equation: \[ \lambda = \frac{h}{p} \] Where: - \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J·s}\)), - \(p\) is the momentum of the particle. Since both the proton and the electron have the same kinetic energy (\(K.E.\)), we can equate their kinetic energies. The kinetic energy is related to momentum as: \[ K.E. = \frac{p^2}{2m} \] Where \(m\) is the mass of the particle and \(p\) is its momentum. Rearranging this equation to solve for momentum, we get: \[ p = \sqrt{2mK.E.} \] Now, since the proton and the electron have the same kinetic energy, their momentum will depend on their masses. The mass of the proton (\(m_p\)) is much greater than the mass of the electron (\(m_e\)), so the momentum of the proton will be greater than that of the electron. This means that the de Broglie wavelength will be inversely proportional to the momentum.
Thus, for the same kinetic energy, the electron, having a smaller mass, will have a smaller momentum and, therefore, a larger de Broglie wavelength compared to the proton.
In summary, the electron has a greater de Broglie wavelength because it has a smaller mass and therefore smaller momentum for the same kinetic energy.