Question

An electron, an alpha particle, a proton and a neutron have same de-Broglie wavelength. What would be the order of their kinetic energies ?

• A. $K.E_e>K.E_p>K.E_n>K.E_{\alpha }$
• B. $K.E_{\alpha }>K.E_n>K.E_p>K.E_e$
• C. $K.E_p=K.E_n=K.E_{\alpha }=K.E_e$
• D. $K.E_p>K.E_n>K.E_{\alpha }>K.E_e$

Answer 1

The correct option is A $K.E_e>K.E_p>K.E_n>K.E_{\alpha }$

Since $\lambda$ is same for all particles, they will also have same momentum.
$\because (p=\frac{h}{\lambda })$

Now, Kinetic Energy is given by, $K.E=\frac{p^2}{2m}$

As $p$ is same for all, therefore $K.E$ will be inversely proportional to mass of the particle.

Masses of the particles are in the following order.

$M_{\alpha }>M_n>M_p>M_e$

$\therefore$ Their $K.E$ will be in the following order.

$K.E_e>K.E_p>K.E_n>K.E_{\alpha }$

Hence, $\left(A\right)$ is the correct answer.

Caution: Such question should only be evaluated based on mathematical information only. Key concept: Kinetic energy and momentum are related as follows, $K.E=\frac{1}{2}M{v}^2=\frac{1}{2}\frac{M^2{v}^2}{M}$ $=\frac{p^2}{2M}(\because p=Mv)$