Question

6. Light with a frequency of 1.61×10151.61×1015 Hz ejects electrons from the surface of zinc, which has a work function of 4.33 eV. What is the minimum de Broglie wavelength of the ejected electrons?
nm

Answer 1

here,

the frequency of light , f = 1.61 * 10^15 Hz

work function , W = 4.33 eV = 4.33 * 1.6 * 10^-19 J

the kinetic energy of ejected electrons , Ke = h * f - W

KE = 6.64 * 10^-34 * 1.61 * 10^15 - 4.33 * 1.6 * 10^-19 J

KE = 3.76 * 10^-19 J

the minimum debroglie wavelength , lamda = h/sqrt(2 * m * KE)

lamda = 6.634 * 10^-34 /sqrt(2 * 9.1 * 10^-31 * 3.76 * 10^-19) m

lamda = 8.02 * 10^-10 m = 0.802 nm