An X - ray tube produces a continuous spectrum of radiation with its shortest wavelength of `45xx10^(-2)Å`. The maximum energy of a photon in the radiation in eV is `(h = 6.62xx10^(-34)Js, c=3xx10^(8)ms^(-1))`

Options

A. 27500
B. 22500
C. 17500
D. 12500

Answer

To find the maximum energy of a photon produced by the X-ray tube, we can use the formula derived from Planck's equation:

\[ E = \frac{hc}{\lambda} \]

Where:
- \( E \) is the energy of the photon in joules,
- \( h \) is Planck's constant (\( 6.62 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^{8} \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.

### Step 1: Convert the wavelength from angstroms to meters
The given wavelength is \( 45 \times 10^{-2} \, \text{Å} \). To convert angstroms to meters, we use the conversion factor \( 1 \, \text{Å} = 10^{-10} \, \text{m} \).

\[ \lambda = 45 \times 10^{-2} \, \text{Å} \times 10^{-10} \, \text{m/Å} = 45 \times 10^{-12} \, \text{m} = 4.5 \times 10^{-11} \, \text{m} \]

### Step 2: Substitute the values into the energy formula
Now we can substitute the values of \( h \), \( c \), and \( \lambda \) into the energy formula.

\[ E = \frac{(6.62 \times 10^{-34} \, \text{Js}) \times (3 \times 10^{8} \, \text{m/s})}{4.5 \times 10^{-11} \, \text{m}} \]

### Step 3: Calculate the energy in joules
Calculating the numerator:

\[ E = \frac{(6.62 \times 3) \times 10^{-34 + 8}}{4.5 \times 10^{-11}} \]

Calculating \( 6.62 \times 3 = 19.86 \):

\[ E = \frac{19.86 \times 10^{-26}}{4.5 \times 10^{-11}} = \frac{19.86}{4.5} \times 10^{-26 + 11} \]

Calculating \( \frac{19.86}{4.5} \approx 4.41 \):

\[ E \approx 4.41 \times 10^{-15} \, \text{J} \]

### Step 4: Convert energy from joules to electron volts
To convert joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \).

\[ E \text{ (in eV)} = \frac{E \text{ (in J)}}{1.6 \times 10^{-19}} \]

Substituting the value of \( E \):

\[ E \text{ (in eV)} = \frac{4.41 \times 10^{-15}}{1.6 \times 10^{-19}} \approx 27500 \, \text{eV} \]

### Final Answer:
The maximum energy of a photon in the radiation is approximately \( 27500 \, \text{eV} \).

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