For the setup illustrated in the figure, y = 1.0 mm , d = 0.24 mm , and D = 1.2 m . The work function of the material of the emitter is 2.2 eV . Determine the stopping potential V required to halt the photocurrent.

Step-by-step solution

From the given figure, the distance from the central bright fringe to the first dark fringe in the interference pattern is y . The position of the first dark fringe is given by: y = D 2d Rearranging for the wavelength of the light source: = 2yd D Substituting the given values y = 1.0 mm = 10⁻³ m , d = 0.24 mm = 0.24 10⁻³ m , and D = 1.2 m : = 2 10⁻³ 0.24 10⁻³ 1.2 = 0.4 10⁻⁶ m = 400 nm The energy of the incident photons is: E = hc = 1240 eV nm 400 nm = 3.1 eV Using Einstein's photoelectric equation, the stopping pot