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 will analyze the given information and apply the relevant formulas.

### Step 1: Understand the relationship between potential difference and wavelength
In a Coolidge tube, the minimum wavelength (\( \lambda_{min} \)) of the emitted X-rays is given by the formula:
\[ \lambda_{min} = \frac{hc}{eV} \]
where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( e \) is the charge of an electron,
- \( V \) is the potential difference.

### Step 2: Calculate the minimum wavelengths for both potential differences
1. **Initial potential difference (\( V_1 = 24.8 \, \text{kV} \)):**
\[ \lambda_{min1} = \frac{hc}{eV_1} = \frac{12.4 \, \text{kV} \cdot \text{Å}}{24.8 \, \text{kV}} = \frac{12.4}{24.8} \, \text{Å} \]

2. **Final potential difference (\( V_2 = 49.6 \, \text{kV} \)):**
\[ \lambda_{min2} = \frac{hc}{eV_2} = \frac{12.4 \, \text{kV} \cdot \text{Å}}{49.6 \, \text{kV}} = \frac{12.4}{49.6} \, \text{Å} \]

### Step 3: Establish the relationship between the K-alpha line wavelength and the minimum wavelength
According to the problem, the difference between the K-alpha wavelength (\( \lambda_k \)) and the minimum wavelength becomes two times the initial K-alpha wavelength:
\[ \lambda_k - \lambda_{min1} = 2 \times \lambda_k - \lambda_{min2} \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ \lambda_k - \lambda_{min1} = 2\lambda_k - \lambda_{min2} \]
\[ \lambda_{min2} - \lambda_{min1} = \lambda_k \]
Thus, we can express \( \lambda_k \) as:
\[ \lambda_k = \lambda_{min2} - \lambda_{min1} \]

### Step 5: Substituting the values
Substituting the expressions for \( \lambda_{min1} \) and \( \lambda_{min2} \):
\[ \lambda_k = \left(\frac{12.4}{49.6}\right) - \left(\frac{12.4}{24.8}\right) \]

### Step 6: Finding a common denominator and simplifying
Finding a common denominator (which is \( 49.6 \times 24.8 \)):
\[ \lambda_k = \frac{12.4 \cdot 24.8 - 12.4 \cdot 49.6}{49.6 \cdot 24.8} \]
\[ = \frac{12.4(24.8 - 49.6)}{49.6 \cdot 24.8} \]
\[ = \frac{12.4 \cdot (-24.8)}{49.6 \cdot 24.8} \]

### Step 7: Final calculation
Calculating gives:
\[ \lambda_k = \frac{-12.4}{49.6} \]
This value can be simplified to find the wavelength of the K-alpha line.

### Conclusion
After performing the calculations, we find the initial wavelength of the K-alpha line.