Question

The following diagram shows the kinetic energy of the ejected photoelectrons against the energy of incident radiation for two metal surfaces M1 and M2. If the energy of the incident radiation on M1 is equal to the work function of M2, the de Broglie wavelength of the ejected photoelectron is nm

[Given: Mass of electron = 9.11x10−31 kg; Planck’s constant = 6.62x10−34 J s; 1 eV=1.6x10−19 J.]
(round off to two decimal places)

Answer 1

The correct answer is: 1.58 to 1.59 (approx)