Question

Consider two particles of different masses. In which of the following situations the heavier of the two particles has smaller de- Broglie wavelength? The two particles

• A. have a free fall through the same height
• B. move with the same kinetic energy
• C. move with the same linear momentum
• D. move with the same speed

Answer 1

Answer: (A), (B), (D)

move with the same speed

The de-Broglie wavelength associated with any particle is,

$\lambda =\frac{h}{p}=\frac{h}{mv}=\frac{h}{\sqrt{2mKE}}.......\left(1\right)$

If both allowed to fall through same height, velocity attained by them will be,

$v=\sqrt{2gh}$ (is same for both)

$\therefore \lambda \propto \frac{1}{m}$ for same $v$.

Let, $m_1\text{and}{m}_2$ be the masses of the two particles.

If, $m_1>m_2\Rightarrow {\lambda }_1<{\lambda }_2$

From (1) we get,

${\lambda }_1=\frac{h}{\sqrt{2m_{1}KE}}{\lambda }_2=\frac{h}{\sqrt{2m_{2}KE}}$

If ${\lambda }_1$ is more, then $m_1$ will be less and vice-versa.

$\Rightarrow$Both the particles will have the same Kinetic energy $\left(KE\right).$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $(A,B)$ and $\left(D\right)$ is the correct answer.