Question

In the photoelectric effect experiment, a beam of light is shining on a metal surface and the electrons are emitted from the metal. One of the three key findings is that a minimum frequency of light is required for emission of electrons. This minimum frequency is found to be 5.5 × 1014 s-1 for an unknown metal. 4 Briefly describe the other two findings and draw a plot fo the KE of emitted electrons against light frequency. (a) (6 marks) (b) Einstein proposed an equation to explain the photoelectric effect. Write down the equation. (1 mark) Calculate the kinetic energy and speed of an electron emitted at a light frequency of 10.8 x 10s1 (c) (3 marks) (d) Determine the de Broglie wavelength of the emitted electron in part (c) (3 marks) Explain how the fast-moving electron beam can be used as the probing light source in electron microscopes (e) (3 marks)

Answer 1

(a) The other two findings of photoelectric effect may be briefly explained as follows:

1. It has been found that the kinetic energy of the electrons ejected is directly proportional to the frequency of the light    (radiation) incident on the metal. Kinetic energy is however, independent of the intensity of the incident radiation.

2. Number of electrons that are ejected from the metal surface per second is dependent on the intensity of the incident radiation, but independent of frequency of the radiation.

A plot of kinetic energy of emitted electrons against frequency is provided below:

Vo Threshold freapuan Yo 2 tre

(b) The equation proposed by Einstein to explain photoelectric effect is provided below:

Where V- fregueny 멱 incident radiation ),ニ ·theshold freg, uenu 2. 2

(c) I have provided the solution in 2 parts. In first image I have solved to calculate the speed of electron. In next image is given solution for kinetic energy of electron.The images are provided below:

the equahon of photbebct effect 2 We kno -34 ะ 10.8 x16"s m mam ot eedtron frequenty, a 91 x1031 I b

Now We know speed of electron (v) ST V 2 18S x 10 mls K.E- (inatic energ 2 -3i -3 I l D 702.338 X 10 2) K.E. ニ 3.51×1019丁

(d) The solution is provided in the image below:

(d) From de Br영lie Egnaton de brog lie wavelenght 7 N S momentum .. man, o{electron X lxio 39 Cas Calaulatud) -34 T114 一10