Question

Revlew Millikan's Photoelectric Experiment Robert A. Mlkan (1868 1953). although best known for his "oil-drop experiment," which measured the charge of an electron, also perfomed pioneering research on the photoelectric effect. In experiments on lithium, for example, Millikan observed a maximum kinetic energy of 0.550 eV when electrons were ejected with 433.9-nm light. When light of 253.5 m was used, he observed a maximum kinetic energy of 2.57 eV.

Part A What is the work function,W, for lithium, as determined from Milikan's results? Express your answer to three significant figures and include appropriate units.
Part B What maximum kinetic energy do you expet illikan found when he used light with a wavelength of 362.4 TIm? Express your answer to three significant figures and include appropriate units Value Units

Answer 1

A.) Work function = 2.3 eV

B.) Max. K.E observed = 1.1 eV

Step-by-step explanation:

A.) Millikan observed a maximum kinetic energy of 0.550 eV when electrons were ejected with 433.9-nm light. When light of 253.5 m was used, he observed a maximum kinetic energy of 2.57 eV.

work function (f) is the minimum energy required to remove an electron from the surface of the material.

hf = Ø + K.E (maximum)

Where

h = Plank constant 6.63 x 10-34 J s

Ø = work function

hc/λ = Ø + K.E (max)

(6.63×10^-34 × 3×10^8)/433.9×10^-9 = Ø + 0.550 × 1.6×10^-19

4.58×10^-19 = Ø + 8.8×10^-20

Ø = 4.58×10^-19 - 8.8×10^-20

Ø = 3.7 × 10^-19 J

Converting Joule to eV

Ø = 3.7 × 10^-19/1.6×10^-19

Ø = 2.3 eV

B.) When light of wavelength 362.4 m is used

The maximum K.E observed = incident light K.E - (the work function).

Incident K.E = hf = hc/λ

Incident K.E =

(6.63×10^-34 × 3×10^8)/362.4

Incident K.E = 5.5 × 10^-28J

Let's convert joule to eV

Incident K.E = 5.5×10^-28/1.6×10^-19

Incident K.E = 3.4 × 10^-9

Max. K.E observed = 3.4 - 2.3

Max. K.E observed = 1.1 eV