Answer:
1.76 eV
Explanation:
Maximum kinetic energy of the emitted photo electrons = energy of the photon- work function of the metal
K.E' = (hc/λ)-∅.................. Equation 1
Where K.E' = maximum kinetic energy of the emitted photo electrons, h = Planck's constant, c = speed, λ = wave length, ∅ = work function of the metal.
make ∅ the subject of the equation
∅ = (hc/λ)-K.E'.................. Equation 2
Given: h = 6.63×10⁻³⁴ J.s, c = 3×10⁸ m/s, λ = 400 nm = 400×10⁻⁹ m, K.E' = 1.1 eV = 1.1(1.602×10⁻¹⁹) J = 1.7622×10⁻¹⁹ J
Substitute into equation 2
∅ = [6.63×10⁻³⁴(3×10⁸)/400×10⁻⁹ ]-1.7622×10⁻¹⁹
∅ = (4.973×10⁻¹⁹)-(1.7622×10⁻¹⁹)
∅ = 3.21×10⁻¹⁹ J.
The maximum kinetic energy of the photo electrons when the wave length is 330 nm is
K.E' = [6.63×10⁻³⁴( 3×10⁸ )/330×10⁻⁹]-(3.21×10⁻¹⁹)
K.E' = (6.03×10⁻¹⁹)-(3.21×10⁻¹⁹)
K.E' = 2.82×10⁻¹⁹ J
K.E' = 2.82×10⁻¹⁹/1.602×10⁻¹⁹
K.E' = 1.76 eV
