In a photoelectric experiment. light is incident on a metal surface Electrons ar

Question

In a photoelectric experiment. light is incident on a metal surface Electrons are ejected from the surface, producing a current in a circuit A reverse potential is applied in the circuit and adjusted until the current drops to zero That potential at which the current drops to zero is called the stopping potential The data obtained for a range of frequencies are graphed below For a frequency of light that has a stopping potential of 3 volts, what is the maximum kinetic energy of the ejected photoelectrons? From the graph and the value of the electron charge, determine an experimental value for Planck's constant From the graph determine the work function for the metal On the axes above, draw the expected graph for a different metal surface with a threshold frequency of 6.0 times 10 14 hertz.

Explanation / Answer

   From thegraph      cut offfrequency      f0   =   5* 1014   Hz    a.   For stoppingpotential   V0   =   3V          Maximumenergy ofphotoelectrons      kmax   =   e* V0      J                                                                                        =   V0   eV                                                                                        =   3.0   eV                                                                                        =   V0   eV                                                                                        =   3.0   eV    b.   For stopping potentialof 0 V, threshold frequency is 5 *1014   Hz           Photo electricequation is          kmax   =   h* f   -          at stoppingpotential,   V0   =   0V,   kmax   =   0,       =>   0   =   h*f0   -   3   eV    h   =   3 eV /f0          =   3* 1.6 * 10-19   J / 5 *1014   Hz          =   9.60* 10-34   J/s          =   9.60* 10-34   J/s    c.   Workfunction         =   h* f0                                              =   9.60* 10-34 * 5 * 1014                                              =   4.80* 10-19   J                                              =   4.80* 10-19 / 1.6 * 10-19                                              =   3.0   eV                                              =   3.0   eV   d.   The graph will be a lineparallel to first line, but it will intercept the x axis at 6 *1014   Hz.

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