Question

Two particles A and B of same mass have their total energies $E_A$ and $E_B$ in the ratio $E_A : E_B = 1 : 2$. Their potential energies $U_A$ and $U_B$ are in the ratio $U_A : U_B = 1 : 2$. If $\lambda_A$ and $\lambda_B$ are their de Broglie wavelengths, then $\lambda_A : \lambda_B$ is

• A. $1 : 2$
• B. $2 : 1$
• C. $1 : \sqrt{2}$
• D. $\sqrt{2} : 1$
• E. $1 : 1$

Hint:

When kinetic energy doubles, the de Broglie wavelength decreases by a factor of $\sqrt{2}$. Always ensure you use kinetic energy, not total energy, in the wavelength formula.

Answer 1

The Correct Option is D

Concept:
The total energy $E$ of a particle is given by $E = K + U$, where $K$ is kinetic energy and $U$ is potential energy. The de Broglie wavelength is related to kinetic energy by $\lambda = \frac{h}{\sqrt{2mK}}$.

Step 1: Determine the ratio of Kinetic Energies ($K$).
Given the ratios: \[ E_A = \frac{1}{2} E_B \quad \text{and} \quad U_A = \frac{1}{2} U_B \] \[ K_A = E_A - U_A \] \[ K_B = E_B - U_B = 2E_A - 2U_A = 2(E_A - U_A) \] Therefore, $K_B = 2K_A$, or the ratio $K_A : K_B = 1 : 2$.

Step 2: Calculate the ratio of de Broglie wavelengths.
Since both particles have the same mass: \[ \lambda \propto \frac{1}{\sqrt{K}} \] \[ \frac{\lambda_A}{\lambda_B} = \sqrt{\frac{K_B}{K_A}} = \sqrt{\frac{2}{1}} = \frac{\sqrt{2}}{1} \] Thus, $\lambda_A : \lambda_B = \sqrt{2} : 1$.