Question

An electron, an $\alpha$ particle and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength?

Answer 1

Kinetic energy is given by $E=\frac{1}{2}m{v}^2$

Same kinetic energy implies that electron has highest velocity (as mass is lowest).

Assume mass of electron as 1 unit. Now, mass of proton approx 1836 and mass of alpha particle is 4 times that is 7344.

Now calculate velocity in each case

For electron $v=\sqrt{2E}$, proton $v=\sqrt{\frac{2E}{1836}}$ and alpha particle $v=\sqrt{\frac{2E}{7344}}$

Also de-Broglie wavelength is given by $\lambda =\frac{h}{mv}$

Now calculate momentum in each case

For electron $mv=\sqrt{2E}\times 1$ proton $mv=\sqrt{\frac{2E}{1836}}\times 1836$ and alpha particle $mv=\sqrt{\frac{2E}{7344}}\times 7344$

Clearly electron has least momentum. Hence it has largest de-Broglie wavelength. And Alpha particle has least wavelength