Question

6 pts 15. Light of three different wavelengths, 250 nm, 400 nm, and 615 nm, is shined on a metal surface. The observations for each wavelength, labeled A, B, and C, follow: Observation A: No photoelectrons were observed Observation B: Photoelectrons with a kinetic energy of 240 kJ/mol were observed Observation C: Photoelectrons with a kinetic energy of 59 kJ/mol were observed Which observation corresponds to which wavelength of light? Justify your choices. Observation A: wavelength=_6015 nm Observation B: wavelength=_ 250 nm Observation C: wavelength = 400 m

Answer 1

Answer: Observation A : Wavelength = 615 mm Observation B : Wavelength = 2500m Observation' c : Wavelength : 400nm Justification for the observations: Observation A: Photselentric effect will be observed when light of sufficient energy falls on a metal surface. The radiation that shines on the metal surface should have a minimum frequency called threshold frequency. If the radiation used has frequency lower than threshold frequency, photo electric effect will not take place. As wavelength and frequency are inversely related, among the 3 wavelengths used, radiation of 615nm has the lowest frequency and hence lowest energy. For the metal used for the experiment this radiation has frequency lower than the threshold frequency and therefore no photoelectric effect was observed. Observation B and C: Kinetic energy of the ejented photselections = half o where his Planck's constant, 2 frequency of radiation used and of work function of the metal. Since the same metal is used & is samed in observations B and C. From the equation (1) it is clear that radiation of higher frequency (ie: lower wavelength) will emit photoelektrons of higher kinetic energy. k. E. of ejuted elections in Observation B is higher and it should be from radiation of lower wavelength ii 250mm. kE of photselections in observation C has lower energy and it should the from radiation of higher wavelength e 400nm.