Answer:
b) 497 nm
Explanation:
Given:
Work function, ϕ = 2.50 eV
Stopping Potential, V₀ = 1.00 eV
Charge of electron, e = 1.6 x 10⁻¹⁹ C
Speed of light, c = 3 x 10⁸ m/s
Planck's constant, h = 6.63 × 10⁻³⁴ Js = 4.14 × 10⁻¹⁵ eVs
Einstein photoelectric equation is given by:
[tex]K.E_{max}[/tex] = E - ϕ ----- (1)
[tex]K.E_{max}[/tex] is the maximum kinetic energy
E is the energy of the absorbed photons: E = hf
ϕ is the work function of the surface: ϕ = hf₀
The potential difference required to back all ejected electrons is called the Stopping Potential (V₀)
Stopping Potential (V₀) [tex]= \frac{K.E. _{max}}{e}[/tex]
[tex]K.E. _{max} = eV_{o}[/tex] -----(2)
Substituting (2) into (1)
eV₀ = E - ϕ
1.6 x 10⁻¹⁹ x 1 = E - 2.50
E = 1.6 x 10⁻¹⁹ + 2.50
E = 2.50 eV
But E = hf = hc/λ
λ [tex]= \frac{hc}{E}[/tex]
λ = (4.14 × 10⁻¹⁵ × 3 x 10⁸) / 2.50
λ = 1240 × 10⁻⁹ / 2.50
λ = 496.8 × 10⁻⁹ m
λ = 497 × 10⁻⁹ m (approximately)
λ = 497 nm
