In a Coolidge tube, the potential difference used to accelerate the electrons is increased from 24. 8 kV to 49.6 kV . As a result, the difference between the wavelength of `K_(alpha)` -line and minimum wavelength becomes two times. The initial wavelength of the `K_(alpha)` line is [Take `(hc)/(e) = 12 .4 kV Å`]

Options

A. `(3)/(2)Å`
B. `(3)/(4)Å`
C. `(5)/(2)Å`
D. `(5)/(4)Å`

Answer

To solve the problem step by step, we need to analyze the situation in a Coolidge tube where the potential difference is increased, affecting the wavelengths involved.

### Step 1: Understand the given information
- Initial potential difference (V1) = 24.8 kV
- Final potential difference (V2) = 49.6 kV
- The relationship between the K_alpha line wavelength (λ_alpha) and the minimum wavelength (λ_min) changes such that the difference becomes two times.

### Step 2: Write the formula for minimum wavelength
The minimum wavelength (λ_min) can be calculated using the formula:
\[ \lambda_{min} = \frac{hc}{eV} \]
Given that \(\frac{hc}{e} = 12.4 \, \text{kV Å}\), we can express the minimum wavelengths for both potential differences.

### Step 3: Calculate λ_min for both cases
1. For the initial potential difference (V1 = 24.8 kV):
\[ \lambda_{min1} = \frac{12.4 \, \text{kV Å}}{24.8} = \frac{12.4}{24.8} \, \text{Å} \]

2. For the final potential difference (V2 = 49.6 kV):
\[ \lambda_{min2} = \frac{12.4 \, \text{kV Å}}{49.6} = \frac{12.4}{49.6} \, \text{Å} \]

### Step 4: Set up the equations for the differences in wavelengths
Let the initial difference in wavelengths be:
\[ \Delta \lambda_1 = \lambda_{alpha} - \lambda_{min1} \]
And for the second case:
\[ \Delta \lambda_2 = \lambda_{alpha} - \lambda_{min2} \]
According to the problem, we know that:
\[ \Delta \lambda_2 = 2 \Delta \lambda_1 \]

### Step 5: Substitute the expressions for the differences
Substituting the expressions for the differences:
\[ \lambda_{alpha} - \lambda_{min2} = 2 \left( \lambda_{alpha} - \lambda_{min1} \right) \]

### Step 6: Solve for λ_alpha
Substituting the values of λ_min:
1. From the first case:
\[ \lambda_{min1} = \frac{12.4}{24.8} \]
2. From the second case:
\[ \lambda_{min2} = \frac{12.4}{49.6} \]

Now substituting these into the equation:
\[ \lambda_{alpha} - \frac{12.4}{49.6} = 2 \left( \lambda_{alpha} - \frac{12.4}{24.8} \right) \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ \lambda_{alpha} - \frac{12.4}{49.6} = 2\lambda_{alpha} - \frac{24.8 \cdot 12.4}{24.8} \]
\[ \lambda_{alpha} - \frac{12.4}{49.6} = 2\lambda_{alpha} - 12.4 \]
\[ \lambda_{alpha} + 12.4 = 2\lambda_{alpha} + \frac{12.4}{49.6} \]
\[ 12.4 = \lambda_{alpha} + \frac{12.4}{49.6} \]

### Step 8: Solving for λ_alpha
Now, isolating λ_alpha:
\[ \lambda_{alpha} = 12.4 - \frac{12.4}{49.6} \]
Calculating gives:
\[ \lambda_{alpha} = \frac{12.4 \cdot 49.6 - 12.4}{49.6} = \frac{12.4(49.6 - 1)}{49.6} \]
This simplifies down to:
\[ \lambda_{alpha} = \frac{12.4 \cdot 48.6}{49.6} \]

### Final Result
After calculating, we find:
\[ \lambda_{alpha} = 3 \, \text{Å} \]