Question:
A metal surface is exposed to a beam of light with a wavelength of 500 nm. The work function of the metal is 3.2 eV. If the energy of a single photon of light is 2.5 eV, determine whether photoelectric effect will occur and state the maximum kinetic energy of the emitted electrons.
Answer:
To determine whether photoelectric effect will occur, we need to compare the energy of the incident photon with the work function of the metal.
Given: Wavelength of light, λ = 500 nm = 500 × 10^(-9) m Energy of a single photon of light, E = 2.5 eV = 2.5 × 1.6 × 10^(-19) J Work function of the metal, Φ = 3.2 eV = 3.2 × 1.6 × 10^(-19) J
The energy of a photon (E) is related to its wavelength (λ) using the equation: E = (hc) / λ
Where: h = Planck's constant = 6.63 × 10^(-34) J·s c = speed of light = 3 × 10^8 m/s
Substituting the given values into the equation, we can calculate the energy of the photon:
E = (6.63 × 10^(-34) J·s × 3 × 10^8 m/s) / (500 × 10^(-9) m) = 3.978 × 10^(-19) J
Comparing the energy of the incident photon with the work function of the metal:
In this case, E = 3.978 × 10^(-19) J and Φ = 3.2 × 1.6 × 10^(-19) J, so photoelectric effect will occur.
The maximum kinetic energy of the emitted electrons can be determined using the equation: K.E. = E - Φ
Substituting the given values into the equation, we can calculate the maximum kinetic energy of the emitted electrons:
K.E. = 3.978 × 10^(-19) J - (3.2 × 1.6 × 10^(-19) J) = 0.554 × 10^(-19) J
Therefore, the maximum kinetic energy of the emitted electrons is 0.554 × 10^(-19) J.