An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at `0.45Ã…`.
The maximum energy of a photon in the radiation is

Options

A. 30.4 ke V
B. 27.6 ke V
C. 15.2 ke V
D. 12.8 ke V

Explanation

To find the maximum energy of a photon produced by an X-ray tube with a given short wavelength, we can use the formula that relates energy and wavelength:

### Step-by-Step Solution:

1. **Identify the formula**: The energy \( E \) of a photon can be expressed in terms of its wavelength \( \lambda \) using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.

2. **Convert the wavelength**: The given wavelength is \( 0.45 \, \text{Å} \) (angstroms). We need to convert this to meters:
\[ 0.45 \, \text{Å} = 0.45 \times 10^{-10} \, \text{m} \]

3. **Substitute the values into the formula**:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{0.45 \times 10^{-10} \, \text{m}} \]

4. **Calculate the energy**:
- First, calculate the numerator:
\[ 6.626 \times 10^{-34} \times 3 \times 10^8 = 1.9878 \times 10^{-25} \, \text{Jm} \]
- Now divide by the wavelength:
\[ E = \frac{1.9878 \times 10^{-25}}{0.45 \times 10^{-10}} = 4.417 \times 10^{-15} \, \text{J} \]

5. **Convert energy from Joules to electron volts**: To convert Joules to electron volts, use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E = \frac{4.417 \times 10^{-15} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 27.6 \times 10^3 \, \text{eV} = 27.6 \, \text{keV} \]

6. **Final Answer**: The maximum energy of the photon in the radiation is:
\[ E \approx 27.6 \, \text{keV} \]