Question

In a photoelectric experiment, you shine light onto an electrode and record a current of 25 μA. When you apply +500 mV to the electrode, the current drops to 19 μA. What is the stopping potential magnitude in V?

Answer 1

Solution: Let initially, when cathode potential is V, mV then photoelectric current is 25uA. When anode is given potential +500 mV then the flow of the current is 19uA and let anode is given potential +V, the current is reduced to zero. This potential V, is the stopping potential when applied to the anode there is no current. Since the current I . Vor V = kl where k is a constant. Then equations for above situations are: V =kx25uA V-500 = kx19 A ...... (2) V;-V, = kx0uA ...... (3) From equation (1) and (2), we have kx25uA + 500 m V = kx19 ÞA=k=. -. Then from 6uA equation (1) and (3), we have V. = kx 25uA = kr.- -x25uA = -2083m V = -2.08V. 6uA Thus the stopping potential is -2.08V.