Question

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

Answer 1

Hint- In order to solve this question, we will use the relation between kinetic energy and momentum of a particle and the De Broglie wavelength formula and then proceed accordingly. After that we will see which one has a larger wavelength by comparing the energy interlinked with momentum of the both particles electron and proton.

Complete step-by-step answer:
In the following equations, p is the magnitude of the relativistic linear momentum, E is the total kinetic energy, m is the invariant mass (also called rest mass), h is Planck's constant, $\lambda$ is the de Broglie wavelength.

De Broglie wavelength formula $\lambda = \dfrac{h}{p}...................\left( 1 \right)$
Kinetic energy is related to momentum as $E = \dfrac{{{p^2}}}{{2m}}................\left( 2 \right)$

From equation (2)
$E = \dfrac{{{p^2}}}{{2m}} \\\ {p^2} = 2mE \\\ p = \sqrt {2mE} ................\left( 3 \right) \\\$

Using equation (1) and equation (3)

$$\lambda = \dfrac{h}{p} \\\ \lambda = \dfrac{h}{{\sqrt {2mE} }} \\\ \lambda \propto \dfrac{1}{m} \\\$$

Since $\lambda$ is inversely proportional to the mass of the electron or proton. Therefore under our assumption of equal kinetic energy for the electron and proton, the de Broglie wavelength of the proton is shorter than for the electron, because the denominator in the last equation is larger for the proton i.e mass of proton is greater than that of electron.

Note- De Broglie equation states that a matter can act as waves much like light and radiation which also behave as waves and particles. The formula also states that an electron beam will also be diffracted like a light beam. In non-relativistic scenarios the heavier the particle, the shorter is the De Broglie wavelength. But in relativistic scenarios, the de Broglie wavelength is nearly independent of the mass of a particle.