Question

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

• A. $k^2 : 1$
• B. $1 : k^2$
• C. $k : 1$
• D. $1 : k$
• E. $1 : 1$

Hint:

If both the kinetic energy and potential energy follow the same ratio, the total energy must also follow that identical ratio.

Answer 1

The Correct Option is B

Concept:
The de Broglie wavelength is $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$, where $K$ is the kinetic energy. Total energy $E = K + U$.

Step 1: Find the ratio of Kinetic Energies ($K$).
Since $\lambda \propto \frac{1}{\sqrt{K}}$ (masses are same): \[ \frac{\lambda_A}{\lambda_B} = \sqrt{\frac{K_B}{K_A}} \implies \frac{k}{1} = \sqrt{\frac{K_B}{K_A}} \implies \frac{K_B}{K_A} = k^2 \implies \frac{K_A}{K_B} = \frac{1}{k^2} \]

Step 2: Find the ratio of Total Energies ($E$).
Given $U_A : U_B = 1 : k^2$, which is the same as the kinetic energy ratio $K_A : K_B = 1 : k^2$. \[ E_A = K_A + U_A \] \[ E_B = K_B + U_B = k^2 K_A + k^2 U_A = k^2(K_A + U_A) \] Therefore, the ratio: \[ \frac{E_A}{E_B} = \frac{K_A + U_A}{k^2(K_A + U_A)} = \frac{1}{k^2} \]