Question

A photon of energy \(10\) units ejects a photo-electron from a metal surface whose work function is \(\Phi _{0}\). If an electron(e) having maximum kinetic energy enters into a uniform magnetic field of induction(B) in a direction perpendicular to the field and describes a circular path of radius R, then radius R is {\(m=\)mass of electron}

• A. \(R=\frac{\sqrt{2m(15-\Phi _{0})}}{eB}\)
• B. \(R=\frac{\sqrt{m(10-\Phi _{0})}}{eB}\)
• C. \(R=\frac{\sqrt{2m(10-\Phi _{0})}}{eB}\)
• D. \(R=\frac{\sqrt{2m(10-\Phi _{0})}}{5eB}\)

Hint:

Find the maximum kinetic energy from Einstein's equation, convert it to momentum, then use the radius of a charged particle in a magnetic field.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The maximum kinetic energy of the photoelectron is \(K_{max}=h\nu-\Phi_0\).
The photon energy is 10 units, so \(K_{max}=10-\Phi_0\).

Step 2: Momentum
\(K=\dfrac{p^{2}}{2m}\), so \(p=\sqrt{2mK_{max}}=\sqrt{2m(10-\Phi_0)}\).

Step 3: Circular Path
For a charge moving perpendicular to the field, \(R=\dfrac{mv}{eB}=\dfrac{p}{eB}\).

Step 4: Calculation
\(R=\dfrac{\sqrt{2m(10-\Phi_0)}}{eB}\), which is option (C).

Final Answer:
The radius equals root of 2m times (10 minus the work function), divided by eB. \[ \boxed{\text{(C) }R=\dfrac{\sqrt{2m(10-\Phi _{0})}}{eB}} \]