Two particles A and B of same mass have their de - Broglie wavelength in the ratio `X_(A):X_(B)=K:1`. Their potential energies `U_(A):U_(B)=1:K^(2)`. The ratio of their total energies `E_(A):E_(B)` is

Options

A. `K^(2):1`
B. `1:K^(2)`
C. `K:1`
D. `1:K`

Answer

To solve the problem, we need to find the ratio of the total energies \( E_A : E_B \) of two particles A and B, given their de Broglie wavelengths and potential energies.

### Step-by-step Solution:

1. **Understanding the Given Ratios**:
- The de Broglie wavelengths are in the ratio \( \lambda_A : \lambda_B = K : 1 \).
- The potential energies are in the ratio \( U_A : U_B = 1 : K^2 \).

2. **Expressing the Kinetic Energy**:
- The de Broglie wavelength is given by the formula:
\[ \lambda = \frac{h}{p} = \frac{h}{mv} \]
where \( p \) is the momentum, \( m \) is the mass, and \( v \) is the velocity of the particle.
- Since both particles have the same mass, we can relate their velocities to their wavelengths:
\[ \lambda_A = \frac{h}{mv_A} \quad \text{and} \quad \lambda_B = \frac{h}{mv_B} \]
- From the ratio \( \lambda_A : \lambda_B = K : 1 \), we can express:
\[ \frac{v_A}{v_B} = \frac{1}{K} \]

3. **Finding Kinetic Energies**:
- The kinetic energy \( K \) of a particle is given by:
\[ KE = \frac{1}{2}mv^2 \]
- Therefore, the kinetic energies can be expressed as:
\[ KE_A = \frac{1}{2}m v_A^2 \quad \text{and} \quad KE_B = \frac{1}{2}m v_B^2 \]
- Using the ratio of velocities:
\[ v_A = \frac{v_B}{K} \implies KE_A = \frac{1}{2}m \left(\frac{v_B}{K}\right)^2 = \frac{1}{2}m \frac{v_B^2}{K^2} = \frac{KE_B}{K^2} \]

4. **Expressing Total Energies**:
- The total energy \( E \) of each particle is the sum of its kinetic and potential energy:
\[ E_A = KE_A + U_A \quad \text{and} \quad E_B = KE_B + U_B \]
- Substituting the expressions for kinetic and potential energies:
\[ E_A = \frac{KE_B}{K^2} + U_A \quad \text{and} \quad E_B = KE_B + U_B \]
- Given \( U_A : U_B = 1 : K^2 \), we can express \( U_A \) and \( U_B \):
\[ U_A = U_B \cdot \frac{1}{K^2} \]

5. **Finding the Final Ratio**:
- Now substituting \( U_A \) in the expression for \( E_A \):
\[ E_A = \frac{KE_B}{K^2} + \frac{U_B}{K^2} \]
- Therefore:
\[ E_A = \frac{1}{K^2}(KE_B + U_B) = \frac{1}{K^2}E_B \]
- Thus, the ratio of total energies is:
\[ \frac{E_A}{E_B} = \frac{1}{K^2} \]

### Final Answer:
The ratio of their total energies \( E_A : E_B = 1 : K^2 \).