The short-wavelength limit shifts by 26 pm when the operating voltage in an X-ray tube is increased to 1.5 times the original value. What was the original value of the operating voltage?

Explanation

To solve the problem step by step, we will use the relationship between the wavelength of X-rays emitted from an X-ray tube and the operating voltage.

### Step 1: Understand the relationship between wavelength and voltage
The wavelength (\( \lambda \)) of the X-rays emitted from an X-ray tube is inversely proportional to the operating voltage (\( V \)). This relationship can be expressed as:
\[ \lambda \cdot V = \frac{hc}{e} \]
where:
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( e \) is the charge of an electron (\( 1.6 \times 10^{-19} \, \text{C} \)).

### Step 2: Define the initial and final conditions
Let:
- \( V \) be the original operating voltage,
- \( \lambda \) be the original wavelength,
- The new voltage after increase is \( V_2 = 1.5V \),
- The new wavelength after the shift is \( \lambda_2 = \lambda - 26 \times 10^{-12} \, \text{m} \).

### Step 3: Set up the equations
Using the relationship established in Step 1, we can write:
\[ \lambda \cdot V = \frac{hc}{e} \quad \text{(1)} \]
\[ \lambda_2 \cdot V_2 = \frac{hc}{e} \quad \text{(2)} \]

From equation (1):
\[ \lambda = \frac{hc}{eV} \]

From equation (2):
\[ \lambda_2 = \frac{hc}{eV_2} \]

### Step 4: Substitute the new conditions into the equations
Substituting \( V_2 = 1.5V \) and \( \lambda_2 = \lambda - 26 \times 10^{-12} \):
\[ \lambda - 26 \times 10^{-12} = \frac{hc}{e(1.5V)} \]

### Step 5: Equate the two expressions for \( \lambda \)
Now we can equate the two expressions for \( \lambda \):
\[ \frac{hc}{eV} - 26 \times 10^{-12} = \frac{hc}{e(1.5V)} \]

### Step 6: Solve for \( V \)
Rearranging gives:
\[ \frac{hc}{eV} - \frac{hc}{e(1.5V)} = 26 \times 10^{-12} \]

Factoring out \( \frac{hc}{e} \):
\[ \frac{hc}{e} \left( \frac{1}{V} - \frac{1}{1.5V} \right) = 26 \times 10^{-12} \]
\[ \frac{hc}{e} \left( \frac{1.5 - 1}{1.5V} \right) = 26 \times 10^{-12} \]
\[ \frac{hc}{e} \left( \frac{0.5}{1.5V} \right) = 26 \times 10^{-12} \]
\[ \frac{hc}{3eV} = 26 \times 10^{-12} \]

Now, substituting the values of \( h \), \( c \), and \( e \):
\[ \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{3(1.6 \times 10^{-19})V} = 26 \times 10^{-12} \]

### Step 7: Calculate \( V \)
Calculating the left side:
\[ \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{3(1.6 \times 10^{-19})} = \frac{6.63 \times 10^{-34} \times 10^8}{1.6 \times 10^{-19}} = \frac{6.63 \times 10^{-26}}{1.6 \times 10^{-19}} = 4.14 \times 10^{-7} \]

Setting it equal to \( 26 \times 10^{-12} \):
\[ \frac{4.14 \times 10^{-7}}{V} = 26 \times 10^{-12} \]
\[ V = \frac{4.14 \times 10^{-7}}{26 \times 10^{-12}} = 15.92 \times 10^{4} \, \text{V} \approx 16 \, \text{kV} \]

### Final Answer
The original value of the operating voltage \( V \) is approximately **16 kV**.