Answer:
13. a) In Einstein's photoelectric equation, hf represents the energy of a single photon of the electromagnetic radiation, and mvmax² represents the maximum kinetic energy of the emitted photoelectrons.
b) i. To calculate the threshold frequency, we can use the equation c = fλ, where c is the speed of light, f is the frequency of the electromagnetic radiation, and λ is the wavelength:
c = fλ
f = c/λ
f = (3.00 x 10^8 m/s)/(6.5 x 10^-7 m)
f = 4.62 x 10^14 Hz
Therefore, the threshold frequency is 4.62 x 10^14 Hz.
ii. We can use the equation hf = Φ + KEmax, where hf is the energy of a single photon, Φ is the work function energy of the metal, and KEmax is the maximum kinetic energy of the emitted photoelectrons. We know that the wavelength of the incident electromagnetic radiation is 6.5 x 10^-7 m, which corresponds to a frequency of f = 4.62 x 10^14 Hz (as calculated in part i). We also know that this wavelength is the maximum for which photoelectrons are released, which means that the energy of the photons is equal to the work function energy:
hf = Φ
Substituting the values for h and f, we get:
(6.63 x 10^-34 J s)(4.62 x 10^14 Hz) = Φ
Φ = 1.93 x 10^-19 J
Converting this to electronvolts (eV), we get:
Φ = (1.93 x 10^-19 J)/(1.60 x 10^-19 J/eV)
Φ = 1.21 eV
Therefore, the work function energy of the metal is 1.9 eV.
c) If the intensity of the incident light is doubled, the rate of release of photoelectrons will also double. This is because the intensity of light is directly proportional to the number of photons incident on the metal surface per unit time. Each photon can cause the emission of one photoelectron, so doubling the number of photons will double the number of emitted photoelectrons per unit time. However, the kinetic energy of the emitted photoelectrons will not change, since this is determined only by the frequency (and therefore the energy) of the incident photons, and not by their intensity.