Lesson Plan 38

November 17, 2017 | Author: suhailielias | Category: Photoelectric Effect, Light, Photon, Solid State Engineering, Physical Universe

STPM...

Description

Lesson Plan Lesson: Concept of Light Quantisation

Aim: Lesson objectives At the end of the lesson, student should be able to •

Use the equation E = hf for a photon

Explain the meaning of work function and threshold frequency

Use Einstein’s equation for photoelectric effect. hf = W +

Explain the meaning of stopping potential and use the equation eVs =

1 2 mv 2 1 2 mv 2

Assumed Prior Knowledge Students should be familiar with •

Understand thermionic emission of electrons from a hot cathode

Understand the quantum theory on light as being made up of photons

c = fλ where c is the velocity of light, f is the frequency of light, and

λ is the wavelength of light Activity Descriptions Activity 1 • Students use the equation E = ht to determine the energy of a photon. -19 • Students learn 1 eV = 1.60 x 10 J Activity 2 • Students learn the phenomenon of photoelectric effect • Students lean the meaning of work function and threshold frequency

1

Activity 3 1 2 • Students use the Einstein’s equation for photoelectric effect, hf = W + mv 2 • For maximum kinetic energy, Kmax = hf − Wo Kmax = hf − hfo Activity 4 1 2 • Students learn and understand the meaning of stopping potential and use the equation eVs = mv

2

Development of Lesson No. 1.

Steps

Set induction

Strategy

Teacher shows video clip of elevator door system that uses a non visible light beam and a photocell as a safety measure. Teacher asks students to explain the effect of blocking the light beam.

2.

Activity

Teacher displays activities of the lesson on the screen one by one, pausing in between to give further explanation and posing questions. Students are invited to try out the worked examples in the activities.

3.

Worksheet

Teacher distributes the printed worksheets to students

Resources

LCD projector and screen. Laptop computer and Courseware LCD projector and screen. Laptop computer and Courseware

Printed worksheet

Students are asked to answer the first two questions. Students are invited to present their answers. Teacher discusses the answer with the students. 4.

Extension

Students are encouraged to surf the Net for further information. Students are asked to try out more problems on the topic from books.

Suggested websites Recommended references.

2

Worksheet Answers Question 1 (a) The energy of one photon, E = hf = =

hc

λ 6.63 × 10 −34 × 3.00 × 10 8 325 × 10 −9 = 6.12 x 10-19 J

(b) Work function of the metal W = 2.70 eV = 2.70×1.6 × 10-19 J = 4.32 x 10-19 J Since energy of one photon is greater than the work function of the metal, electrons will be ejected. (c) Kmax = h f − W = 6.12 x 10-19 J − 4.32 x 10-19 J = 1.80 x 10-19 J Question 2 The energy of one photon,

(6.63 × 10 −34 )(3.00 × 10 8 ) E= = = 3.98 × 10 −19 J −9 λ 500 × 10 hc

Power P = n × energy of one photon

n=

1000 = 2.51× 10 21 per metre square per second −19 3.98 ×10

Question 3 (a) Work function is the minimum amount of energy needed to free an electron from the surface of a metal. Different metals have different work functions. (b) The threshold wavelength is the longest wavelength of radiation that will cause an electron to be ejected from a material. Any wavelength longer than this does not have enough energy. (c) Stopping potential is the potential difference needed to bring an electron of maximum kinetic energy to rest.

3

Question 4 (a) An electron emerging from the metal with zero kinetic energy absorbs just enough energy to escape from the metal, so hf = W + 0 From the graph, the threshold frequency is the intercept on the x-axis,

1 2 mvmax = 0. 2

where

Intercept on x-axis fo = 4.5 x 1014 Hz W = h fo = (6.63 x 10-34) (4.5 x 1014) = 3.0 x 10-19 J (b)

hc

1 2 = W + mvmax λ 2 −34 8 (6.63 × 10 )(3.0 × 10 ) 1 2 = 3.0 x 10 -19 + mvmax −9 2 300 × 10 1 2 mvmax = 3.6 x 10 -19 J 2 hf =

Question 5 (i) Work function W = h fo

= (6.63 x 10-34 )(5.5 x 1014) J =

(6.63 x 10 -34 )(5.5 x 1014 ) eV 1.6 ×10 −19

1 eV = 1.6 × 10-19 J

= 2.3 eV (ii) Kmax = h f − W =

=

hc

λ

−W

6.63×10 −34 × 3.0 × 10 8 − 3.65 ×10 −19 −9 300 ×10

= 2.980 ×10 −19 J

4

1 2 mvmax = 2.980 ×10 −19 2 2 × 2.980 ×10 −19 9.1 × 10 −31 = 8.1× 10 5 m s −1

vmax =

Question 6 Wavelength of blue light λ = 450 nm. The energy emitted in one second = 60 J and P = n h f where n is the number of photons per second

n=

Pλ 60 × 450 ×10 −9 = =1.4 ×10 20. −34 8 hc 6.63× 10 × 3.00 ×10

This value is an over estimation because some of the energy is transformed into heat energy. Question 7 Stopping Potential Vs /V

4.0 3.0

x x

2.0 x

1.0 x 0.0

0

2

4

6

8

10

12

14

Frequency of incident light f / (1014 Hz)

-1.0

-2.0

5

eVs = hf − W Vs =

h W f − e e

The gradient of the graph is

h e

and the intercept on the stopping potential axis is at −

W e

W = -1.0 V e

W =1.6 ×10-19 J

h ∆Vo 3.0 − (−1.0) = = = 4.0 × 10 −15 14 ∆f (10.0 − 0) × 10 e h = 4.0 × 10 −15 ×1.60 × 10-19 = 6.4× 10-34 J s Question 8

W 4.3 × 10 −19 = 6.49 × 1014 Hz = −34 h 6.63 × 10 c 3.0 × 108 Frequency of the incident light f = = = 9.23 × 1014 Hz −9 λ 325 × 10

The threshold frequency

fo =

The stopping potential

h( f − f o) e 6.63 × 10 − 34 × (9.23 × 1014 − 6.49 × 1014 ) = 1.6 × 10 −19 =1.14 V

Vs =