If the potential difference V applied to the coolidge tube is doubled, then the cut off wavelength

Options

A. Is doubled
B. Is halved
C. Remains unchanged
D. Is quadrupled

Answer

To solve the problem of how the cutoff wavelength changes when the potential difference \( V \) applied to the Coolidge tube is doubled, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between energy and wavelength**:
The energy \( E \) of the emitted X-rays is related to the wavelength \( \lambda \) by the equation:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light.

2. **Relate energy to potential difference**:
When a charge \( Q \) is accelerated through a potential difference \( V \), the energy gained by the charge is given by:
\[ E = QV \]
For the electrons in the Coolidge tube, the charge \( Q \) is constant.

3. **Set the two energy expressions equal**:
Since both expressions represent the energy of the emitted X-rays, we can equate them:
\[ QV = \frac{hc}{\lambda} \]

4. **Express the cutoff wavelength in terms of potential**:
Rearranging the equation gives us:
\[ \lambda = \frac{hc}{QV} \]
This shows that the wavelength \( \lambda \) is inversely proportional to the potential difference \( V \).

5. **Analyze the effect of doubling the potential**:
If the potential difference \( V \) is doubled (i.e., \( V \rightarrow 2V \)), we can express the new wavelength \( \lambda_2 \) as:
\[ \lambda_2 = \frac{hc}{Q(2V)} = \frac{hc}{2QV} = \frac{\lambda}{2} \]

6. **Conclusion**:
Therefore, when the potential difference is doubled, the cutoff wavelength is halved:
\[ \lambda_2 = \frac{\lambda}{2} \]

### Final Answer:
The cutoff wavelength is halved when the potential difference is doubled.