When a photon of energy 18.0 eV strikes the surface of a metal having a work function of 2.0 eV , photoelectrons are emitted. What is the approximate de Broglie wavelength of the fastest emitted photoelectron? (Given: h = 6.63 10⁻³⁴ Js , m_e = 9.1 10⁻³¹ kg , 1 eV = 1.6 10⁻¹⁹ J )

Step-by-step solution

According to Einstein's photoelectric equation, the maximum kinetic energy ( KE_ max ) of the emitted photoelectron is: KE_ max = E_ incident - W₀ KE_ max = 18.0 eV - 2.0 eV = 16.0 eV The de Broglie wavelength ( ) of an electron is given by: = h 2m_e KE For an electron, this can be simplified using the standard relation: = 12.27 V where V is the accelerating potential in volts equivalent to the kinetic energy in eV. Since KE_ max = 16.0 eV , the equivalent potential is V = 16 V . = 12.27 16 = 12.27 4 = 3.0675 3.07