The radiation corresponding to 3 → 2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. The electrons are passed through a magnetic field of 5 × 10⁻⁴ T. Assume that the radius of the largest circular path followed by these electrons is 7 mm, the work function of the metal is : (Mass of electron = 9.1 × 10⁻³¹ kg)

Solution

Physics sir · 4 steps

The largest circle belongs to the fastest electron, so the radius gives you the maximum kinetic energy.

1. Photon energy. For the 3 → 2 transition, E = 13.6(1/2² - 1/3²) = 13.6 × 5/36 = 1.89 eV.
2. Radius to momentum. In a magnetic field r = mv/eB, so a bigger r means a faster electron. The largest radius belongs to the fastest electron, the one with maximum kinetic energy. Its momentum is p = eBr.

   Kₘₐₓ = p²/2m = (eBr)²/2m

3. Put in numbers. p = 1.6 × 10⁻¹⁹ × 5 × 10⁻⁴ × 7 × 10⁻³ = 5.6 × 10⁻²⁵ kg m s⁻¹. Then Kₘₐₓ = (5.6 × 10⁻²⁵)²/(2 × 9.1 × 10⁻³¹) = 1.72 × 10⁻¹⁹ J. Dividing by 1.6 × 10⁻¹⁹ gives 1.08 eV.
4. So. φ = E - Kₘₐₓ = 1.89 - 1.08 ≈ 0.81 eV. The closest option is 0.82 eV, option (c).

The idea to keep. A charged particle's radius in a magnetic field gives its momentum, p = qBr. Convert that to kinetic energy with p²/2m, then use Einstein's equation φ = hν - Kₘₐₓ.

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