Question

In a demonstration of the photoelectric effect, suppose that a minimum energy of 5.2×10^−19 J (3.2 eV) is required to dislodge an electron from a metal surface.

What is the minimum frequency (and longest wavelength) of radiation for which the detector registers a response?
Express your answer using two significant figures.
f_min = ______________ Hz
λ_max = ______________ nm

Answer 1

f_min = 7.8 × 10^−14 Hz
λ_max = 380 nm

Minimum energy
Φ_0 -> work-function
-> Φ_0 = 5.2 × 10^−19 J (3.2 eV)
Frequency is f_0
Einstein photo electric effect -> E = Φ_0 + K
E -> energy of photon & K -> kinetic energy
E -> minimum & K -> O & h -> Planck's constant
E = Φ_0 + O
h × f = Φ_0
f = Φ_0 / h
f = (5.2 × 10^−19) / (6.63 × 10^−34) = 7.843 × 10^−14
ANSWER -> f = 7.8 × 10^−14 Hz

Speed of photon
c = 3 × 10^8 m/s
wavelength -> λ = c / f
λ = (3 × 10^8) / (7.8 × 10^−14) = 3.846 × 10^21
(1 nm = 10^-9 m)
ANSWER -> λ = 380 nm