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 will follow these steps:

### Step 1: Understand the given ratios
We are given:
- The ratio of total energies: \( E_A : E_B = 1 : 2 \)
- The ratio of potential energies: \( U_A : U_B = 1 : 2 \)

### Step 2: Express total energy in terms of potential and kinetic energy
The total energy \( E \) of an atom can be expressed as:
\[ E = U + K \]
where \( U \) is potential energy and \( K \) is kinetic energy.

For atoms A and B, we can write:
\[ E_A = U_A + K_A \]
\[ E_B = U_B + K_B \]

### Step 3: Set up the equations based on the ratios
From the given ratios, we can express the total energies as:
\[ E_A = U_A + K_A \]
\[ E_B = U_B + K_B \]

Using the ratio \( E_A : E_B = 1 : 2 \), we can write:
\[ 2E_A = E_B \]
Substituting the expressions for \( E_A \) and \( E_B \):
\[ 2(U_A + K_A) = U_B + K_B \]

### Step 4: Substitute the potential energy ratio
From the ratio of potential energies \( U_A : U_B = 1 : 2 \), we can express:
\[ U_B = 2U_A \]

### Step 5: Substitute \( U_B \) into the energy equation
Substituting \( U_B \) into the energy equation gives:
\[ 2(U_A + K_A) = 2U_A + K_B \]

### Step 6: Simplify the equation
Now, we can simplify the equation:
\[ 2U_A + 2K_A = 2U_A + K_B \]
Cancelling \( 2U_A \) from both sides results in:
\[ 2K_A = K_B \]
Thus, we find:
\[ \frac{K_A}{K_B} = \frac{1}{2} \]

### Step 7: Relate de Broglie wavelengths to kinetic energy
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
where \( p \) is momentum. The momentum can also be expressed in terms of kinetic energy \( K \):
\[ p = \sqrt{2mK} \]
Thus, we can express the de Broglie wavelengths for atoms A and B as:
\[ \lambda_A = \frac{h}{\sqrt{2mK_A}} \]
\[ \lambda_B = \frac{h}{\sqrt{2mK_B}} \]

### Step 8: Find the ratio of de Broglie wavelengths
Now, we can find the ratio of the de Broglie wavelengths:
\[ \frac{\lambda_A}{\lambda_B} = \frac{\sqrt{2mK_B}}{\sqrt{2mK_A}} = \sqrt{\frac{K_B}{K_A}} \]

### Step 9: Substitute the kinetic energy ratio
From our earlier result, we have:
\[ \frac{K_B}{K_A} = 2 \]
Thus, we can substitute this into the wavelength ratio:
\[ \frac{\lambda_A}{\lambda_B} = \sqrt{2} \]

### Final Result
Therefore, the ratio of the de Broglie wavelengths is:
\[ \lambda_A : \lambda_B = \sqrt{2} : 1 \]