The maximum mass of photons emitted by an X-ray tube when a voltage of 27 kV is applied to it is given as `N xx 10^(-32) kg`. Find the value of N.

Explanation

To find the maximum mass of photons emitted by an X-ray tube when a voltage of 27 kV is applied, we can follow these steps:

### Step 1: Understand the energy conversion
The energy of the photons emitted from the X-ray tube is derived from the kinetic energy gained by the electrons when they are accelerated through the applied voltage. The energy (E) gained by an electron when accelerated through a potential difference (V) is given by the equation:

\[ E = eV \]

where:
- \( e \) is the charge of the electron (approximately \( 1.6 \times 10^{-19} \) coulombs),
- \( V \) is the voltage (27 kV = \( 27 \times 10^3 \) volts).

### Step 2: Calculate the energy in joules
Substituting the values into the equation:

\[ E = (1.6 \times 10^{-19} \, \text{C}) \times (27 \times 10^3 \, \text{V}) \]

Calculating this gives:

\[ E = 1.6 \times 27 \times 10^{-16} \]
\[ E = 43.2 \times 10^{-16} \]
\[ E = 4.32 \times 10^{-15} \, \text{J} \]

### Step 3: Relate energy to mass
According to Einstein’s mass-energy equivalence principle, the energy of a photon can also be expressed in terms of its mass (m) using the equation:

\[ E = mc^2 \]

where:
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)).

### Step 4: Solve for mass
Rearranging the equation to solve for mass gives:

\[ m = \frac{E}{c^2} \]

Substituting the energy calculated in step 2:

\[ m = \frac{4.32 \times 10^{-15}}{(3 \times 10^8)^2} \]

Calculating \( c^2 \):

\[ c^2 = (3 \times 10^8)^2 = 9 \times 10^{16} \]

Now substituting this value:

\[ m = \frac{4.32 \times 10^{-15}}{9 \times 10^{16}} \]
\[ m = 4.8 \times 10^{-32} \, \text{kg} \]

### Step 5: Identify the value of N
From the expression \( m = N \times 10^{-32} \, \text{kg} \), we can see that:

\[ N = 4.8 \]

### Final Answer
The value of \( N \) is **4.8**.

---