Question

5. Solar panels have become a source of “Green” power. Those that directly convert sunlight into electricity are called photo-voltaic cells and use the photo-electric effect to generate their electricity. a. Classical wave theory stated that by increasing the intensity (amount) of light, that each electron would have more kinetic energy, however experiments showed that increasing the intensity increased the number of electrons released, but not the amount of energy each electron had. Why is this so? b. A photon of blue light with a wavelength of 420nm hits the panel. How much energy does the photon have, in electron Volts? c. If the solar panel emits electrons with an of 1.270 eV, show that the work function is 1.682 eV. d. If the work function is 1.682 eV/photon, what is this in Joules/photon? e. What is the cut off frequency?

Answer 1

(a) According to the classical wave theory, the intensity (I) of light (which is a transverse electro magnetgic wave according to Maxwells equations) is directly proportional to the square of its amplitude (A).  

                                                                         $I\propto \left | A \right |^{2}$

Therefore, increasing intensity means increasing the wave amplitude. Hence, it is conceivable that if incident light intensity is increased, electrons in the metal will oscillate more violently and will be ejected with higher kinetic energy.

However, experiments does not support this explanation. Later, in the year of 1905, photoelectric effect has been well explained by Einstein considering light made up of discrete tiny partcles called photons. Each photon has energy E = h$\nu$, where $\nu$ is the frequency of the incident light. Intensity of light is defined as the number of photons incident on per unit area of the suface per unit time. Hence, increasing light intensity does not increase the energy of each photon. Energy of each photon is dependent only on its frequency $\nu$ . So, kinetic energy of emitted photoelectrons remains constant as long as incident photon frequency is kept constant. This can also be understood from the Einstein famous photoelectric effect equation

                                                      $(K.E)_{max}=h\nu -\phi _{0}$

From the above equation we can see that for any metal ($\phi _{0}$ is a constant). Hence, kinetic energy of emitted photoelectrons only depent of the frequency of the incident photons and not on the intensity of the incident photon beam.

(6) a = 420 mm = 420X10-9m E = Who a 6.63 x 10-2 2-34 х 3x108 Joule 420 X10-9 17 FDE 0.04735X107 Joule -. E = -19 0.04735 X10 1: 6 X10-19 ev 11 0.02959 x102 E 11 =D 2.959 ev (0) Photo eletric eff ut equation is, (K. Emax = hood. 52 1.270 = 2.959 - 60 1.689 ev d) % = 1.689 ev. = d. (1.689 x 1.6 210– ) joules Po 2. 7024x10-19 goales. le emitted At cut-off frequency, photo current on becomes Zero. (K.E) mex = 0, hNo-4 = 0 photo eletrons