Find the cut off walelength for the continuous X - rays coming
form an X-ray tube operating at 40 kv.

Explanation

To find the cut-off wavelength for the continuous X-rays coming from an X-ray tube operating at 40 kV, we can follow these steps:

### Step 1: Understand the relationship between energy and wavelength
The energy (E) of a photon is related to its wavelength (λ) by the equation:
\[ E = \frac{hc}{\lambda} \]
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.

### Step 2: Convert the operating voltage to energy
The energy of the electrons accelerated through a potential difference (V) is given by:
\[ E = eV \]
where:
- \( e \) is the charge of an electron (\( 1.6 \times 10^{-19} \, \text{C} \)),
- \( V \) is the voltage in volts.

For a tube operating at 40 kV:
\[ E = 1.6 \times 10^{-19} \, \text{C} \times 40 \times 10^3 \, \text{V} = 6.4 \times 10^{-15} \, \text{J} \]

### Step 3: Convert energy from joules to electron volts
To convert joules to electron volts, we can use the conversion:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus, the energy in electron volts is:
\[ E = \frac{6.4 \times 10^{-15} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} = 40 \, \text{eV} \]

### Step 4: Calculate the cut-off wavelength
Now, substituting the energy back into the wavelength equation:
\[ \lambda = \frac{hc}{E} \]
We can use \( hc \approx 1242 \, \text{eV} \cdot \text{nm} \):
\[ \lambda = \frac{1242 \, \text{eV} \cdot \text{nm}}{40 \, \text{eV}} = 31.05 \, \text{nm} \]

### Final Answer
The cut-off wavelength for the continuous X-rays coming from an X-ray tube operating at 40 kV is approximately:
\[ \lambda \approx 31.05 \, \text{nm} \]