In an X-ray tube the value of applied potential difference is 81 kV. The emitted radiations may contain wavelength of

Options

A. `2xx10^(-11) m,1.2xx10^(-11)`m
B. `2.1xx10^(-11) m , 1.0xx10^(-11)` m
C. `2.0xx10^(-11)m , 2.5xx10^(-11)`m
D. `1.3xx10^(-11)m , 1.0xx10^(-11)` m

Explanation

To find the emitted radiation wavelengths from an X-ray tube with an applied potential difference of 81 kV, we can use the relationship between energy and wavelength given by the formula:

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

Where:
- \(E\) is the energy of the emitted photon,
- \(h\) 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.

The energy \(E\) can also be expressed in terms of the charge of an electron and the potential difference \(V\):

\[ E = eV \]

Where:
- \(e\) is the charge of an electron (\(1.6 \times 10^{-19} \, \text{C}\)),
- \(V\) is the potential difference in volts.

### Step-by-step Solution:

1. **Convert the Potential Difference to Volts**:
The given potential difference is 81 kV, which is:
\[ V = 81 \times 10^3 \, \text{V} \]

2. **Calculate the Energy of the Electrons**:
Using the formula \(E = eV\):
\[ E = (1.6 \times 10^{-19} \, \text{C})(81 \times 10^3 \, \text{V}) = 1.296 \times 10^{-14} \, \text{J} \]

3. **Use the Energy to Find the Minimum Wavelength**:
Rearranging the energy-wavelength formula gives:
\[ \lambda = \frac{hc}{E} \]
Substituting the values:
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{1.296 \times 10^{-14} \, \text{J}} \]

4. **Calculate the Wavelength**:
\[ \lambda = \frac{1.9878 \times 10^{-25}}{1.296 \times 10^{-14}} \approx 1.53 \times 10^{-11} \, \text{m} \]

5. **Determine Possible Wavelengths**:
The emitted radiation may contain wavelengths that are equal to or greater than the minimum wavelength calculated. Therefore, any wavelength option given in the question must be greater than or equal to \(1.53 \times 10^{-11} \, \text{m}\).

### Conclusion:
The emitted radiation may contain wavelengths that are equal to or greater than \(1.53 \times 10^{-11} \, \text{m}\).