Question

The Photoelectric Effect A Review Constants Periodic Table Electrons are emitted from the surface of a metal when it's exposed to light. This is called the photoelectric effect. Each metal has a certain threshold frequency of light, below which nothing happens. Right at this threshold frequency, an electron is emitted. Above this frequency, the electron is emitted and the extra energy is transferred to the electron The equation for this phenomenon is Here are some data collected on a sample of cesium exposed to various energies of light. Light energy Electron emitted? (eV) Electron KE (eV) no no KE=hy - hw yes 3.87 3.88 3.89 3.90 3.91 *Note that 1 eV (electron volt) = 1.60 x 10-19 J. yes where KE is the kinetic energy of the emitted electron, h=6.63 x 10 5 J .s is Planck's constant, v is the frequency of the light, and is the threshold frequency of the metal. 0.01 yes 0.02 Also, since E=hy, the equation can also be written as KE=E-6 Part A is the binding energy where is the energy of the light and of the electron in the metal. What is the threshold frequency 0 of cesium? Express your answer numerically in hertz.

Answer 1

K.E. = ho ho V = Threshold frequency from 3rd data KE=0 ho = 3. 89 ev Simiotol = 3.89*16*10? J -lg .hv = h - 3.89 X 1.6 x 10'9 J = 3.89 x 1- 6x10-19 blod 663 x 10-34 J = 0.939 x 10'5 Jono 1 = 9.39 x 1014 Hz Hz