The potential difference applied to an X-ray tube is V The ratio of the de Broglie wavelength of electron to the minimum wavlength of X-ray is directrly proportional to

Options

A. V volt
B. `sqrtV`
C. `V^((3)/(2))`
D. `V((7)/(2))`

Explanation

To solve the problem, we need to find the ratio of the de Broglie wavelength of an electron to the minimum wavelength of X-rays, and determine what this ratio is directly proportional to.

### Step-by-step Solution:

1. **Understanding de Broglie Wavelength**:
The de Broglie wavelength (\( \lambda_e \)) of an electron is given by the formula:
\[ \lambda_e = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the electron.

2. **Finding the Momentum of the Electron**:
The momentum \( p \) of the electron can be expressed in terms of its kinetic energy. For an electron accelerated through a potential difference \( V \), the kinetic energy \( KE \) is given by:
\[ KE = eV \]
where \( e \) is the charge of the electron. The relationship between kinetic energy and momentum is:
\[ KE = \frac{p^2}{2m} \]
Therefore, we can express momentum as:
\[ p = \sqrt{2m \cdot KE} = \sqrt{2m \cdot eV} \]

3. **Substituting Momentum into the de Broglie Wavelength**:
Now substituting the expression for momentum into the de Broglie wavelength formula:
\[ \lambda_e = \frac{h}{\sqrt{2m \cdot eV}} \]

4. **Finding the Minimum Wavelength of X-rays**:
The minimum wavelength (\( \lambda_x \)) of X-rays produced in an X-ray tube can be expressed as:
\[ \lambda_x = \frac{hc}{eV} \]
where \( c \) is the speed of light.

5. **Finding the Ratio**:
Now we need to find the ratio of the de Broglie wavelength of the electron to the minimum wavelength of X-rays:
\[ \frac{\lambda_e}{\lambda_x} = \frac{\frac{h}{\sqrt{2m \cdot eV}}}{\frac{hc}{eV}} = \frac{eV}{hc} \cdot \frac{h}{\sqrt{2m \cdot eV}} = \frac{eV \cdot h}{hc \cdot \sqrt{2m \cdot eV}} \]
Simplifying this gives:
\[ \frac{\lambda_e}{\lambda_x} = \frac{1}{c} \cdot \frac{\sqrt{eV}}{\sqrt{2m}} \]

6. **Conclusion**:
From the above expression, we can see that the ratio \( \frac{\lambda_e}{\lambda_x} \) is directly proportional to \( \sqrt{V} \). Therefore, the answer is:
\[ \frac{\lambda_e}{\lambda_x} \propto \sqrt{V} \]

### Final Answer:
The ratio of the de Broglie wavelength of the electron to the minimum wavelength of X-ray is directly proportional to \( \sqrt{V} \).