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 regarding the effect of doubling the potential difference \( V \) applied to a Coolidge tube on the cut-off wavelength, we can follow these steps:

### Step 1: Understand the relationship between energy and potential difference
The energy \( E \) gained by an electron when accelerated through a potential difference \( V \) is given by the equation:
\[ E = eV \]
where \( e \) is the charge of the electron.

### Step 2: Relate energy to wavelength
The energy of a photon can also be expressed in terms of its wavelength \( \lambda \) using the equation:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light.

### Step 3: Set the two expressions for energy equal
Since both expressions represent the same energy, we can set them equal to each other:
\[ eV = \frac{hc}{\lambda} \]

### Step 4: Rearrange the equation to find the cut-off wavelength
Rearranging the equation to solve for the cut-off wavelength \( \lambda \), we get:
\[ \lambda = \frac{hc}{eV} \]

### Step 5: Analyze the effect of doubling the potential difference
If the potential difference \( V \) is doubled (i.e., \( V' = 2V \)), we can substitute \( V' \) into the equation for \( \lambda \):
\[ \lambda' = \frac{hc}{e(2V)} = \frac{hc}{2eV} \]

### Step 6: Compare the new wavelength with the original wavelength
From the above equation, we can see that:
\[ \lambda' = \frac{1}{2} \cdot \frac{hc}{eV} = \frac{1}{2} \lambda \]
This shows that the new cut-off wavelength \( \lambda' \) is half of the original cut-off wavelength \( \lambda \).

### Final Conclusion
Thus, if the potential difference \( V \) applied to the Coolidge tube is doubled, the cut-off wavelength is halved.