When `24.8keV` X- rays strike a material, the photoelectrons emitted from K shell are observed to move in a circle of radius 23 mm in a magnetic field of `2xx10^(-2)T`. The binding energy of K shell electrons is

Options

A. 6.2 keV
B. 5.4 keV
C. 7.4 keV
D. 8.6 keV

Answer

To solve the problem, we need to find the binding energy of the K shell electrons when 24.8 keV X-rays strike a material, and the emitted photoelectrons move in a circular path in a magnetic field.

### Step-by-Step Solution:

1. **Identify the given values:**
- Energy of X-rays (E_x) = 24.8 keV = \( 24.8 \times 10^3 \) eV
- Radius of circular motion (r) = 23 mm = \( 23 \times 10^{-3} \) m
- Magnetic field (B) = \( 2 \times 10^{-2} \) T
- Charge of electron (e) = \( 1.6 \times 10^{-19} \) C
- Mass of electron (m) = \( 9.1 \times 10^{-31} \) kg

2. **Calculate the velocity of the photoelectrons:**
The centripetal force required for circular motion is provided by the magnetic force. The equation is:
\[ \frac{mv^2}{r} = e v B \]
Rearranging gives:
\[ v = \frac{eBr}{m} \]

3. **Substituting the values:**
\[ v = \frac{(1.6 \times 10^{-19} \, \text{C})(2 \times 10^{-2} \, \text{T})(23 \times 10^{-3} \, \text{m})}{9.1 \times 10^{-31} \, \text{kg}} \]

4. **Calculating the velocity:**
\[ v = \frac{(1.6 \times 10^{-19})(2 \times 10^{-2})(23 \times 10^{-3})}{9.1 \times 10^{-31}} = \frac{7.36 \times 10^{-24}}{9.1 \times 10^{-31}} \approx 8.09 \times 10^{6} \, \text{m/s} \]

5. **Calculate the kinetic energy (KE) of the photoelectrons:**
The kinetic energy is given by:
\[ KE = \frac{1}{2} mv^2 \]
Substituting the value of v:
\[ KE = \frac{1}{2} m \left( \frac{eBr}{m} \right)^2 = \frac{1}{2} \frac{e^2 B^2 r^2}{m} \]

6. **Substituting the values into the kinetic energy formula:**
\[ KE = \frac{1}{2} \frac{(1.6 \times 10^{-19})^2 (2 \times 10^{-2})^2 (23 \times 10^{-3})^2}{9.1 \times 10^{-31}} \]

7. **Calculating the kinetic energy:**
\[ KE \approx \frac{1}{2} \frac{(2.56 \times 10^{-38})(4 \times 10^{-4})(5.29 \times 10^{-4})}{9.1 \times 10^{-31}} \approx 2.97 \times 10^{-15} \, \text{J} \]
Converting to eV:
\[ KE \approx \frac{2.97 \times 10^{-15}}{1.6 \times 10^{-19}} \approx 18.36 \, \text{keV} \]

8. **Calculate the binding energy of the K shell:**
The binding energy (BE) can be found using:
\[ BE = E_x - KE \]
\[ BE = 24.8 \, \text{keV} - 18.36 \, \text{keV} = 6.44 \, \text{keV} \]

### Final Answer:
The binding energy of K shell electrons is approximately **6.44 keV**.