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: Formula First
For a charged particle in a magnetic field, $evB=\dfrac{mv^{2}}{R}$, so $R=\dfrac{mv}{eB}$.

Step 2: Speed from Energy
$\dfrac{1}{2}mv^{2}=10-\Phi_0$, so $mv=\sqrt{2m(10-\Phi_0)}$.

Step 3: Result
Substituting, $R=\dfrac{\sqrt{2m(10-\Phi_0)}}{eB}$, which is option (C).

Final Answer:
Substituting the momentum from the kinetic energy gives option (C). \[ \boxed{\text{(C) }R=\dfrac{\sqrt{2m(10-\Phi _{0})}}{eB}} \]