The de-Broglie wavelength of electrons striking the cathode to produce X-rays is equal to `(1)/(100)` of minimum wavelength of X-ray produced. What is the voltage (in volt) across X-ray tube?

Answer

To solve the problem, we need to find the voltage (V) across the X-ray tube given that the de-Broglie wavelength (λ_d) of electrons striking the cathode is equal to \( \frac{1}{100} \) of the minimum wavelength (λ_min) of the X-ray produced.

### Step-by-Step Solution:

1. **Understanding the relationship between de-Broglie wavelength and energy**:
The de-Broglie wavelength of an electron can be expressed as:
\[ \lambda_d = \frac{h}{mv} \]
where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( m \) is the mass of the electron (\( 9.11 \times 10^{-31} \, \text{kg} \)),
- \( v \) is the velocity of the electron.

2. **Relating the minimum wavelength of X-rays to the energy**:
The minimum wavelength of X-rays produced can be related to the energy (E) of the electrons striking the cathode:
\[ \lambda_{min} = \frac{hc}{E} \]
where:
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( E \) is the energy of the electrons, which can also be expressed in terms of voltage (V) as:
\[ E = eV \]
where \( e \) is the charge of an electron (\( 1.6 \times 10^{-19} \, \text{C} \)).

3. **Setting up the relationship**:
Given that:
\[ \lambda_d = \frac{1}{100} \lambda_{min} \]
Substituting the expressions for λ_d and λ_min, we have:
\[ \frac{h}{mv} = \frac{1}{100} \cdot \frac{hc}{eV} \]

4. **Simplifying the equation**:
Canceling \( h \) from both sides:
\[ \frac{1}{mv} = \frac{c}{100eV} \]
Rearranging gives:
\[ mv = \frac{100eV}{c} \]

5. **Expressing velocity in terms of voltage**:
The kinetic energy of the electrons can also be expressed as:
\[ E = \frac{1}{2} mv^2 = eV \]
Thus:
\[ v = \sqrt{\frac{2eV}{m}} \]

6. **Substituting for v**:
Substituting this expression for v into the previous equation:
\[ m\sqrt{\frac{2eV}{m}} = \frac{100eV}{c} \]
Squaring both sides:
\[ m^2 \cdot \frac{2eV}{m} = \left(\frac{100eV}{c}\right)^2 \]
Simplifying gives:
\[ 2meV = \frac{10000e^2V^2}{c^2} \]

7. **Rearranging to find V**:
Rearranging the equation to isolate V:
\[ V = \frac{2mc^2}{10000e} \]

8. **Substituting known values**:
Substituting \( m = 9.11 \times 10^{-31} \, \text{kg} \), \( c = 3 \times 10^8 \, \text{m/s} \), and \( e = 1.6 \times 10^{-19} \, \text{C} \):
\[ V = \frac{2 \times (9.11 \times 10^{-31}) \times (3 \times 10^8)^2}{10000 \times (1.6 \times 10^{-19})} \]

9. **Calculating the voltage**:
Performing the calculations:
\[ V \approx 102.3 \, \text{volts} \]

### Final Answer:
The voltage across the X-ray tube is approximately **102.3 volts**.