An X-ray tube is operated at 1.24 million volt. The shortest wavelength of the produced photon will be :

Options

A. `10^(-3)` nm
B. `10^(-1)` nm
C. `10^(-2)` nm
D. `10^(-4)` nm

Answer

To find the shortest wavelength of the X-ray photon produced in an X-ray tube operated at 1.24 million volts, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Energy of the Photon**:
The energy (E) of a photon can be related to its wavelength (λ) using the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{J s} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.

2. **Relating Energy to Voltage**:
The energy of the electrons accelerated through a potential difference (V) can be expressed in electron volts (eV). Since 1 eV = \( 1.6 \times 10^{-19} \, \text{J} \), the total energy in joules for 1.24 million volts is:
\[ E = 1.24 \times 10^6 \, \text{V} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.984 \times 10^{-13} \, \text{J} \]

3. **Substituting Energy into the Wavelength Equation**:
Now, substituting the energy into the wavelength equation:
\[ \lambda = \frac{hc}{E} \]

4. **Calculating λ**:
Plugging in the values of \( h \), \( c \), and \( E \):
\[ \lambda = \frac{(6.63 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{1.984 \times 10^{-13} \, \text{J}} \]
\[ \lambda = \frac{1.989 \times 10^{-25} \, \text{J m}}{1.984 \times 10^{-13} \, \text{J}} \approx 1.003 \times 10^{-12} \, \text{m} \]

5. **Converting to Angstroms**:
To convert meters to angstroms (1 angstrom = \( 10^{-10} \, \text{m} \)):
\[ \lambda \approx 1.003 \times 10^{-12} \, \text{m} = 10.03 \, \text{pm} = 0.1003 \, \text{nm} = 1.003 \, \text{Å} \]

6. **Final Answer**:
The shortest wavelength of the produced photon will be approximately:
\[ \lambda \approx 0.1 \, \text{nm} \text{ or } 1.003 \, \text{Å} \]