If the work function of a metal is 6.63 × 10⁻¹⁹J, the maximum wavelength of the photon required to remove a photoelectron from the metal is ____________ nm. (Nearest integer)
[Given : h = 6.63 × 10⁻³⁴ J s, and c = 3 × 10⁸ m s⁻¹]

Solution

Chemistry ma’am · 3 steps

The longest wavelength that works is the one whose photon energy just equals the work function.

1. Threshold condition. A photon removes an electron only if E = hc/λ is at least the work function φ. Energy falls as wavelength rises, so the maximum wavelength is the one where E = φ exactly.

   λₘₐₓ = hc/φ

2. Put in numbers. λₘₐₓ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(6.63 × 10⁻¹⁹) m. The 6.63 cancels, leaving 3 × 10⁻³⁴⁺⁸⁺¹⁹ = 3 × 10⁻⁷ m.
3. So. Since 1 nm = 10⁻⁹ m, 3 × 10⁻⁷ m = 300 nm. The answer is 300.

The common slip: 3000. Converting 3 × 10⁻⁷ m to ångström instead of nanometre gives 3000, not 300.

The idea to keep. Threshold wavelength comes from λ₀ = hc/φ. Any longer wavelength carries too little energy per photon, however bright the light.

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