The voltage applied to an X-ray tube is `18 kV`. The maximum mass of photon emitted by the X-ray tube will be

Options

A. `2 xx 10^(-13) kg`
B. `3.2 xx 10^(-36) kg`
C. `3.2 xx 10^(-32) kg`
D. `9.1 xx 10^(-31) kg`

Explanation

To find the maximum mass of a photon emitted by an X-ray tube when a voltage of 18 kV is applied, we can follow these steps:

### Step 1: Understand the relationship between energy and mass
The energy of a photon can be related to its mass using Einstein's equation:
\[ E = mc^2 \]
where:
- \( E \) is the energy,
- \( m \) is the mass,
- \( c \) is the speed of light in a vacuum (\( c \approx 3 \times 10^8 \, \text{m/s} \)).

### Step 2: Determine the maximum energy of the photon
The maximum energy of the photon emitted from the X-ray tube is equal to the energy gained by an electron when it is accelerated through a potential difference \( V \):
\[ E = eV \]
where:
- \( e \) is the charge of the electron (\( e \approx 1.6 \times 10^{-19} \, \text{C} \)),
- \( V \) is the voltage applied (in this case, \( V = 18 \, \text{kV} = 18 \times 10^3 \, \text{V} = 1.8 \times 10^4 \, \text{V} \)).

### Step 3: Calculate the energy of the photon
Substituting the values into the equation:
\[ E = eV = (1.6 \times 10^{-19} \, \text{C})(1.8 \times 10^4 \, \text{V}) \]
Calculating this gives:
\[ E = 2.88 \times 10^{-15} \, \text{J} \]

### Step 4: Relate energy to mass
Now, we can relate the energy to mass using \( E = mc^2 \):
\[ m = \frac{E}{c^2} \]
Substituting the values:
\[ m = \frac{2.88 \times 10^{-15} \, \text{J}}{(3 \times 10^8 \, \text{m/s})^2} \]

### Step 5: Calculate the mass
Calculating \( c^2 \):
\[ c^2 = (3 \times 10^8)^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \]
Now substituting back to find \( m \):
\[ m = \frac{2.88 \times 10^{-15}}{9 \times 10^{16}} \]
\[ m = 3.2 \times 10^{-32} \, \text{kg} \]

### Final Answer
Thus, the maximum mass of the photon emitted by the X-ray tube is:
\[ m \approx 3.2 \times 10^{-32} \, \text{kg} \]

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