The photo electric work function of
potassium surface is 2.2ev. When
ultra violet light of wavelength 3200A
falls on potassium surface , calculate
the energy of the most energetic
photon emitted. What is the
threshold frequency for potassium.
Answers
Given info : The photo electric work function of
potassium surface is 2.2ev. When
ultra violet light of wavelength 3200A
falls on potassium surface.
To find : the energy of the most energetic photon emitted and the threshold frequency for potassium.
solution : work function of potassium surface, Φ = 2.2 eV
wavelength of ultra violet light falls on potassium surface, λ = 3200 A°
so, incident energy, E = hc/λ = 12400/(λ in A°) eV
= 12400/3200
= 3.875 eV
from photoelectric effect equation,
K.E_max = E - Φ
= 3.875 eV - 2.2 eV
= 1.675 eV
so, the energy of the most energetic photon emitted is 1.675 eV.
threshold frequency, f = work function/Plank's constant
= (2.2 × 1.6 × 10^-19)/(6.63 × 10^-34) Hz
= 0.531 × 10¹⁵ Hz
= 5.31 × 10¹⁴ Hz
Therefore the threshold frequency for potassium is 5.31 × 10¹⁴ Hz.