One increasing the operating voltage in a x-ray tube by `1.5` times, the shortest wavelength decreases by `26 "pm"`. Find the original value of operating voltage.

Options

A. `~~10 kV`
B. `~~16 kV`
C. `~~50 kV`
D. `~~75 kV`

Answer

To solve the problem, we need to find the original value of the operating voltage (V_initial) in an X-ray tube given that increasing the voltage by 1.5 times decreases the shortest wavelength by 26 picometers.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The relationship between the operating voltage (V) and the shortest wavelength (λ) in an X-ray tube is given by the equation:
\[ E \cdot V = \frac{hc}{\lambda} \]
where \(E\) is the charge of an electron, \(h\) is Planck's constant, and \(c\) is the speed of light.

2. **Setting Up the Equations**:
From the relationship above, we can say that:
\[ V \cdot \lambda = \text{constant} \]
This means that if the voltage changes, the wavelength will also change accordingly.

3. **Defining the Variables**:
Let:
- \(V_{\text{initial}} = V\)
- \(V_{\text{final}} = 1.5V\)
- \(λ_{\text{initial}} = λ\)
- \(λ_{\text{final}} = λ - 26 \text{ pm}\)

4. **Using the Relationship**:
According to the relationship:
\[ V_{\text{initial}} \cdot λ_{\text{initial}} = V_{\text{final}} \cdot λ_{\text{final}} \]
Substituting \(V_{\text{final}} = 1.5V_{\text{initial}}\) and \(λ_{\text{final}} = λ_{\text{initial}} - 26 \text{ pm}\):
\[ V \cdot λ = 1.5V \cdot (λ - 26) \]

5. **Simplifying the Equation**:
Dividing both sides by \(V\) (assuming \(V \neq 0\)):
\[ λ = 1.5(λ - 26) \]
Expanding the right side:
\[ λ = 1.5λ - 39 \]

6. **Rearranging the Equation**:
Bringing all terms involving \(λ\) to one side:
\[ 39 = 1.5λ - λ \]
Simplifying gives:
\[ 39 = 0.5λ \]
Therefore,
\[ λ = 78 \text{ pm} \]

7. **Converting Wavelength to Meters**:
Convert picometers to meters:
\[ λ = 78 \times 10^{-12} \text{ m} \]

8. **Calculating the Original Voltage**:
Now we can use the equation \(E \cdot V = \frac{hc}{λ}\) to find \(V_{\text{initial}}\):
\[ V_{\text{initial}} = \frac{hc}{λ} \]
Substituting \(h = 6.626 \times 10^{-34} \text{ J s}\), \(c = 3 \times 10^8 \text{ m/s}\), and \(λ = 78 \times 10^{-12} \text{ m}\):
\[ V_{\text{initial}} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{78 \times 10^{-12}} \]

9. **Calculating the Result**:
Performing the calculation:
\[ V_{\text{initial}} = \frac{1.9878 \times 10^{-25}}{78 \times 10^{-12}} \approx 2541.67 \text{ volts} \]

10. **Final Result**:
Therefore, the original operating voltage is approximately:
\[ V_{\text{initial}} \approx 2541.67 \text{ volts} \text{ or } 2.54 \text{ kV} \]