When `0.50` Å X-ray strike a material , the photoelectron from the `K` shell are observed to move in a circle of radius `2.3 nm` in a magnetic field of `2 xx 10^(-2)` tesle acting perpendicular to direct of emission of photoelectron . What is the binding elergy of `k-shell` electron?

Explanation

To find the binding energy of the K-shell electron, we will follow these steps:

### Step 1: Determine the velocity of the photoelectron
The magnetic force acting on the photoelectron provides the centripetal force required for circular motion. The magnetic force can be expressed as:
\[ F = QVB \]
where:
- \( Q \) is the charge of the electron (\( Q = e = 1.6 \times 10^{-19} \, \text{C} \)),
- \( V \) is the velocity of the electron,
...