The operating potential in an x-ray tube is increased by 2%. The percentage change in the cut off wavelength is

Options

A. 1% increase
B. 2% increase
C. 2% decrease
D. 1% decrease

Explanation

To solve the problem of finding the percentage change in the cut-off wavelength when the operating potential in an X-ray tube is increased by 2%, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**: The cut-off wavelength (\( \lambda_{min} \)) in an X-ray tube is given by the formula:
\[ \lambda_{min} = \frac{h \cdot c}{e \cdot V} \]
where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( e \) is the charge of an electron,
- \( V \) is the operating potential.

2. **Differentiate the Expression**: To find the change in the cut-off wavelength with respect to a change in potential, we can differentiate the equation:
\[ \frac{\Delta \lambda_{min}}{\lambda_{min}} = -\frac{\Delta V}{V} \]
This indicates that an increase in potential (\( V \)) will result in a decrease in the cut-off wavelength (\( \lambda_{min} \)).

3. **Calculate the Percentage Change**: Given that the operating potential is increased by 2%, we can express this as:
\[ \frac{\Delta V}{V} \times 100 = 2\% \]
Therefore, substituting this into the differentiated expression:
\[ \frac{\Delta \lambda_{min}}{\lambda_{min}} \times 100 = -\Delta V/V \times 100 = -2\% \]

4. **Interpret the Result**: The negative sign indicates that the cut-off wavelength decreases. Thus, the percentage change in the cut-off wavelength is:
\[ \text{Percentage change in } \lambda_{min} = -2\% \]

5. **Final Answer**: The percentage change in the cut-off wavelength is a decrease of 2%. Therefore, the correct option is:
- **Option 3: 2% decrease**.