Two electrons in two hydrogen - like atoms A and B 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

Options

A. `1 : 2`
B. `2 : 1`
C. `1:sqrt(2)`
D. `sqrt(2) : 1`

Answer

To solve the problem, we need to find the ratio of the de-Broglie wavelengths of two electrons in hydrogen-like atoms A and B, given the ratios of their total energies and potential energies.

### Step-by-step Solution:

1. **Understanding the Ratios**:
- Given:
\[ E_A : E_B = 1 : 2 \]
\[ U_A : U_B = 1 : 2 \]
- Let \( E_A = E \) and \( E_B = 2E \).
- Let \( U_A = U \) and \( U_B = 2U \).

2. **Kinetic Energy Relation**:
- The total energy \( E \) of an electron in a hydrogen-like atom can be expressed as:
\[ E = K + U \]
- The potential energy \( U \) is negative, and we know that:
\[ U = -2K \]
- Thus, we can express the total energy in terms of kinetic energy:
\[ E = K + U = K - 2K = -K \]
- Therefore, for atom A:
\[ E_A = -K_A \quad \text{and} \quad E_B = -K_B \]
- From the ratios, we can write:
\[ -K_A = E \quad \text{and} \quad -K_B = 2E \]
- This gives us:
\[ K_A = -E \quad \text{and} \quad K_B = -2E \]

3. **Kinetic Energy Ratio**:
- From the above, we find:
\[ K_A : K_B = E : 2E = 1 : 2 \]

4. **Momentum Relation**:
- The kinetic energy \( K \) is related to momentum \( p \) as:
\[ K = \frac{p^2}{2m} \]
- Thus, we can express momentum in terms of kinetic energy:
\[ p_A = \sqrt{2m K_A} \quad \text{and} \quad p_B = \sqrt{2m K_B} \]
- Therefore, the ratio of momenta is:
\[ \frac{p_A}{p_B} = \frac{\sqrt{2m K_A}}{\sqrt{2m K_B}} = \sqrt{\frac{K_A}{K_B}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

5. **De-Broglie Wavelength**:
- The de-Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
- Thus, the ratio of the de-Broglie wavelengths is:
\[ \frac{\lambda_A}{\lambda_B} = \frac{p_B}{p_A} \]
- Substituting the ratio of momenta:
\[ \frac{\lambda_A}{\lambda_B} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} \]

6. **Final Ratio**:
- Therefore, the ratio of the de-Broglie wavelengths is:
\[ \lambda_A : \lambda_B = \sqrt{2} : 1 \]

### Conclusion:
The final answer is:
\[ \lambda_A : \lambda_B = \sqrt{2} : 1 \]