The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 Aᵒ is 0.42 V. If the threshold frequency is x × 10¹³ /s, where x is _________ (nearest integer).
(Given, speed light = 3 × 10⁸ m/s, Planck's constant = 6.63 × 10⁻³⁴ Js)

Solution

Physics sir · 4 steps

Find the photon energy, take away the stopping-potential energy, and what is left is the work function.

1. Photon energy. With λ = 6630 Å = 6.63 × 10⁻⁷ m, the numbers are chosen so that h and λ cancel neatly: E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(6.63 × 10⁻⁷) = 3 × 10⁻¹⁹ J.
2. Maximum kinetic energy. The stopping potential tells you the fastest electron's kinetic energy: eV₀ = 0.42 × 1.6 × 10⁻¹⁹ = 0.672 × 10⁻¹⁹ J.

   hν₀ = hc/λ - eV₀ = 3 × 10⁻¹⁹ - 0.672 × 10⁻¹⁹ = 2.328 × 10⁻¹⁹ J

3. Threshold frequency. Divide the work function by h: ν₀ = (2.328 × 10⁻¹⁹)/(6.63 × 10⁻³⁴) = 3.51 × 10¹⁴ s⁻¹.
4. So. ν₀ = 35.1 × 10¹³ s⁻¹, so x rounds to 35.

The common slip: 45. Using the frequency of the incident light, c/λ = 4.52 × 10¹⁴ s⁻¹, instead of the threshold frequency gives 45. The energy eV₀ must be subtracted first.

The idea to keep. Einstein's equation hν = hν₀ + eV₀ links everything. Convert the stopping potential to joules with e, subtract, then divide by h to get the threshold frequency.

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