An X-ray tube is opearted at 18 kV. The maximum velocity of electron striking the target is

Options

A. `8xx10^7 m//s`
B. `6xx10^7 m//s`
C. `5xx10^7 m//s`
D. None of these

Explanation

To find the maximum velocity of electrons striking the target in an X-ray tube operated at 18 kV, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between voltage and kinetic energy**:
- When an electron is accelerated through a potential difference (voltage) \( V \), it gains kinetic energy equal to the work done on it by the electric field. This can be expressed as:
\[ KE = eV \]
where \( e \) is the charge of the electron and \( V \) is the voltage.

2. **Express kinetic energy in terms of velocity**:
- The kinetic energy of an electron can also be expressed in terms of its mass \( m \) and velocity \( v \):
\[ KE = \frac{1}{2} mv^2 \]
Here, we are interested in the maximum velocity \( v_{max} \) of the electron.

3. **Set the two expressions for kinetic energy equal**:
- Since both expressions represent the kinetic energy of the electron, we can set them equal to each other:
\[ eV = \frac{1}{2} mv_{max}^2 \]

4. **Rearrange the equation to solve for \( v_{max} \)**:
- Rearranging the equation gives us:
\[ v_{max}^2 = \frac{2eV}{m} \]
- Taking the square root of both sides, we find:
\[ v_{max} = \sqrt{\frac{2eV}{m}} \]

5. **Substitute the known values**:
- The charge of an electron \( e = 1.6 \times 10^{-19} \) C.
- The mass of an electron \( m = 9.1 \times 10^{-31} \) kg.
- The voltage \( V = 18 \) kV = \( 18 \times 10^3 \) V.
- Now substituting these values into the equation:
\[ v_{max} = \sqrt{\frac{2 \times (1.6 \times 10^{-19} \, \text{C}) \times (18 \times 10^3 \, \text{V})}{9.1 \times 10^{-31} \, \text{kg}}} \]

6. **Calculate the value**:
- First, calculate the numerator:
\[ 2 \times (1.6 \times 10^{-19}) \times (18 \times 10^3) = 5.76 \times 10^{-15} \]
- Now divide by the mass of the electron:
\[ \frac{5.76 \times 10^{-15}}{9.1 \times 10^{-31}} \approx 6.32 \times 10^{15} \]
- Finally, take the square root:
\[ v_{max} \approx \sqrt{6.32 \times 10^{15}} \approx 8 \times 10^7 \, \text{m/s} \]

7. **Conclusion**:
- The maximum velocity of electrons striking the target is approximately \( 8 \times 10^7 \, \text{m/s} \).