A metal surface is illuminated by a radiation of wavelength 4500 Aᵒ. The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90° with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximately :

Solution

Physics sir · 3 steps

The circle's radius gives the electron's momentum, so you get its KE and subtract it from the photon energy.

1. Momentum from circle. The magnetic force provides the centripetal force, so evB = mv²/r, which gives mv = eBr. Here p = (1.6 × 10⁻¹⁹)(2 × 10⁻³)(2 × 10⁻³) = 6.4 × 10⁻²⁵ kg m s⁻¹.

   K = p²/2m = (6.4 × 10⁻²⁵)²/(2 × 9.1 × 10⁻³¹) = 2.25 × 10⁻¹⁹ J ≈ 1.41 eV

2. Photon energy. E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(4.5 × 10⁻⁷) = 4.42 × 10⁻¹⁹ J ≈ 2.76 eV.
3. So. By Einstein's equation, φ = E - K = 2.76 - 1.41 ≈ 1.36 eV, which is option (a).

The idea to keep. A charged particle moving perpendicular to B on a circle has p = qBr. Turn that into K = p²/2m, then use φ = hc/λ - K.

Answer from the NTA final answer key. Solution by PrepWiser.