Question

When \(0.5\)\AA X-rays strike a material, the photoelectrons from the K-shell are observed to move in a circular radius \(23\,\text{mm}\) in a magnetic field of \(2\times10^{-2}\,\text{T}\) perpendicular to the direction of emission of photoelectrons. What is the binding energy of K-shell electrons?

• A. \(3.5\,\text{keV}\)
• B. \(6.2\,\text{keV}\)
• C. \(2.9\,\text{keV}\)
• D. \(5.5\,\text{keV}\)

Hint:

Magnetic field gives momentum directly: \[ p=eBr \] Then use photoelectric equation.

Answer 1

The Correct Option is B

Step 1: Momentum of photoelectron: \[ p = eBr \]
Step 2: Kinetic energy: \[ K = \frac{p^2}{2m} = \frac{e^2B^2r^2}{2m} \]
Step 3: Energy of incident photon: \[ E = \frac{hc}{\lambda} \]
Step 4: Binding energy: \[ E_b = E - K \]
Step 5: Substituting values gives: \[ E_b = 6.2\,\text{keV} \]