In a Coolidge tube, the potential difference used to accelerate the electrons is increased from `12.4 kV` to `24.8 kV`. As a result, the difference between the wavelengths of `K_(alpha)`-line and minimum wavelength becomes thrice. The wavelength of the `K_(alpha)` line is `0.25xxKA^(@)`. Find the value of `K`. `(hc)/(e)=(12.4 KVA^(@))`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We are given that the potential difference in a Coolidge tube is increased from 12.4 kV to 24.8 kV, and the difference between the wavelengths of the K-alpha line and the minimum wavelength becomes three times greater. The wavelength of the K-alpha line is given as \( \lambda_{K\alpha} = 0.25 K \) Å. We need to find the value of \( K \).

### Step 2: Write the formula for minimum wavelength
The minimum wavelength \( \lambda_{min} \) for X-rays can be calculated using 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 in volts.

### Step 3: Calculate the minimum wavelength for both potentials
1. For \( V_1 = 12.4 \) kV:
\[ \lambda_{min1} = \frac{hc}{e \cdot 12.4 \times 10^3} \]

2. For \( V_2 = 24.8 \) kV:
\[ \lambda_{min2} = \frac{hc}{e \cdot 24.8 \times 10^3} \]

### Step 4: Write the expressions for the differences in wavelengths
1. The difference in wavelengths when \( V_1 \) is applied:
\[ \Delta_1 = \lambda_{K\alpha} - \lambda_{min1} \]

2. The difference in wavelengths when \( V_2 \) is applied:
\[ \Delta_2 = \lambda_{K\alpha} - \lambda_{min2} \]

### Step 5: Set up the equation based on the problem statement
According to the problem, the difference in wavelengths when the potential is increased becomes three times greater:
\[ \Delta_2 = 3 \Delta_1 \]

### Step 6: Substitute the expressions for \( \Delta_1 \) and \( \Delta_2 \)
Substituting the expressions we derived:
\[ \lambda_{K\alpha} - \lambda_{min2} = 3(\lambda_{K\alpha} - \lambda_{min1}) \]

### Step 7: Substitute the minimum wavelength expressions
Substituting the values of \( \lambda_{min1} \) and \( \lambda_{min2} \):
\[ \lambda_{K\alpha} - \frac{hc}{e \cdot 24.8 \times 10^3} = 3\left(\lambda_{K\alpha} - \frac{hc}{e \cdot 12.4 \times 10^3}\right) \]

### Step 8: Simplify the equation
Rearranging the equation:
\[ \lambda_{K\alpha} - \frac{hc}{e \cdot 24.8 \times 10^3} = 3\lambda_{K\alpha} - \frac{3hc}{e \cdot 12.4 \times 10^3} \]

### Step 9: Solve for \( \lambda_{K\alpha} \)
Combine like terms and solve for \( \lambda_{K\alpha} \):
\[ 2\lambda_{K\alpha} = \frac{hc}{e \cdot 24.8 \times 10^3} - \frac{3hc}{e \cdot 12.4 \times 10^3} \]

### Step 10: Substitute the known value of \( \frac{hc}{e} \)
Given \( \frac{hc}{e} = 12.4 K \) Å:
\[ 2\lambda_{K\alpha} = \frac{12.4 K}{24.8} - \frac{3 \cdot 12.4 K}{12.4} \]

### Step 11: Solve for \( K \)
After simplifying, we find that:
\[ \lambda_{K\alpha} = 1.25 \text{ Å} \]
Substituting back into the equation \( \lambda_{K\alpha} = 0.25 K \):
\[ 0.25 K = 1.25 \implies K = \frac{1.25}{0.25} = 5 \]

### Final Answer
The value of \( K \) is \( 5 \).

---