A potential difference of 30 kV is applied across an X-ray tube. Find the minimum wavelength of X-ray generated .

Options

A. `7.07 xx 10^(-2) "Å"`
B. `4.133 xx 10^(-10) m`
C. `7.07 xx 10^(-10)m`
D. `4.133 xx 10^(-11) m`

Answer

To find the minimum wavelength of X-rays generated when a potential difference of 30 kV is applied across an X-ray tube, we can use the relationship between energy and wavelength. The energy (E) of the X-rays can be calculated using the formula:

\[ E = Q \cdot V \]

Where:
- \( E \) is the energy in joules,
- \( Q \) is the charge of an electron (approximately \( 1.6 \times 10^{-19} \) coulombs),
- \( V \) is the potential difference in volts (30 kV = \( 30,000 \) volts).

1. **Calculate the energy (E):**
\[ E = Q \cdot V = (1.6 \times 10^{-19} \, \text{C}) \cdot (30,000 \, \text{V}) \]
\[ E = 4.8 \times 10^{-15} \, \text{J} \]

2. **Use the energy-wavelength relationship:**
The energy of a photon is also related to its wavelength (\( \lambda \)) by the equation:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^{8} \, \text{m/s} \)).

3. **Rearranging the equation to find the wavelength (\( \lambda \)):**
\[ \lambda = \frac{hc}{E} \]

4. **Substituting the values:**
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{Js}) \cdot (3 \times 10^{8} \, \text{m/s})}{4.8 \times 10^{-15} \, \text{J}} \]

5. **Calculating \( \lambda \):**
\[ \lambda = \frac{1.9878 \times 10^{-25}}{4.8 \times 10^{-15}} \]
\[ \lambda = 4.138 \times 10^{-11} \, \text{m} \]

Thus, the minimum wavelength of the X-ray generated is approximately \( 4.138 \times 10^{-11} \, \text{m} \) or \( 41.38 \, \text{pm} \).