Question:
When \(0.5\,\text{\AA}\) X-rays strike a material, the photoelectrons from the K-shell are observed to move in a circular path of radius \(23\,\text{mm}\) in a magnetic field of \(2 \times 10^{-2}\,\text{T}\), perpendicular to the direction of emission of photoelectrons. What is the binding energy of K-shell electrons?
When \(0.5\,\text{\AA}\) X-rays strike a material, the photoelectrons from the K-shell are observed to move in a circular path of radius \(23\,\text{mm}\) in a magnetic field of \(2 \times 10^{-2}\,\text{T}\), perpendicular to the direction of emission of photoelectrons. What is the binding energy of K-shell electrons?
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Magnetic field gives momentum directly:
\[
p=eBr
\]
Then use photoelectric equation.
Updated On: Apr 2, 2026
- \(3.5\,\text{keV}\)
- \(6.2\,\text{keV}\)
- \(2.9\,\text{keV}\)
- \(5.5\,\text{keV}\)
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The Correct Option is B
Solution and Explanation
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: \[ E_b = 6.2\,\text{keV} \]
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