A light beam of wavelength 500 nm is incident on a metal having work function of 1.25 eV, placed in a magnetic field of intensity B. The electrons emitted perpendicular to the magnetic field B, with maximum kinetic energy are bent into circular are of radius 30 cm. The value of B is ___________ × 10⁻⁷ T.
Given hc = 20 × 10⁻²⁶ J-m, mass of electron = 9 × 10⁻³¹ kg

Solution

Physics sir · 4 steps

Find the electron's maximum kinetic energy first, turn it into momentum, then r = p/eB gives B.

1. Photon energy. E = hc/λ = (20 × 10⁻²⁶)/(500 × 10⁻⁹) = 4 × 10⁻¹⁹ J. Dividing by 1.6 × 10⁻¹⁹ gives 2.5 eV.
2. Maximum kinetic energy. Kₘₐₓ = 2.5 - 1.25 = 1.25 eV = 2 × 10⁻¹⁹ J. Its momentum is p = √2mK = √(2 × 9 × 10⁻³¹ × 2 × 10⁻¹⁹) = √(36 × 10⁻⁵⁰) = 6 × 10⁻²⁵ kg m s⁻¹.

   r = mv/eB = √2mK/eB ⇒ B = √2mK/er

3. Put in numbers. With r = 0.3 m: B = (6 × 10⁻²⁵)/(1.6 × 10⁻¹⁹ × 0.3) = (6 × 10⁻²⁵)/(4.8 × 10⁻²⁰) = 1.25 × 10⁻⁵ T.
4. So. B = 125 × 10⁻⁷ T, so the answer is 125.

The common slip: 1.25. Putting r = 30 (in cm) instead of 0.3 m makes B a hundred times too small and gives 1.25, not 125.

The idea to keep. For a charge circling in a magnetic field, r = √2mK/qB. Get K from Einstein's equation in joules, and keep every length in metres.

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