photoelectric effect + electron diffraction

Question 1

State what is meant by the photoelectrons.

Answer 1

electrons emitted due to light

Question 2

Explain the theoretical vs the experimental view of photoelectrons’ KE.

Answer 2

• theoretically, we observe photoelectrons with a range of different energies
• experimentally, maximum kinetic energy of photoelectrons.

Question 3

State the effect of light intensity on maximum kinetic energy of photoelectrons.

Answer 3

max EK is independent of light intensity.

Question 4

State the two hypotheses’ made by wave theorists about the photoelectric effect.

Answer 4

→ lag time between light source being turned on & photoelectrons being emitted
→ EK max is dependant on light intensity

Question 5

State the effect of increasing light intensity on the photoelectric effect.

Answer 5

no. of photoelectrons emitted per second increases

Question 6

State what determines the maximum kinetic energy of photoelectrons.

Answer 6

frequency.

Question 7

For a given metal plate, when will photoelectrons be emitted?

Answer 7

only if the frequency of the light source is above a certain value

Question 8

For a given metal plate, when will photoelectrons be emitted?

Answer 8

→ frequency > threshold frequency
→ if photon energy > work function

Question 9

State what is meant by the work function.

Answer 9

the minimum energy required to remove an electron from a metal surface Φ

Question 10

What is a photon?

Answer 10

a packet of energy that makes up light

Question 11

State the relationship between photon energy and frequency.

Answer 11

increasing frequency increases photon energy

Question 12

State how to calculate the energy of a photon.

Answer 12

E = hf
→ E is energy in joules
→ h is Planck’s constant in Js
→ f is frequency in hertz

Question 13

State what is meant by threshold frequency.

Answer 13

the minimum frequency of incident electromagnetic radiation required for an electron to overcome the work function

Question 14

Explain how to find the threshold frequency.

Answer 14

1) using E = hf, set E to work function
2) rearrange for f(min) and solve

Question 15

Use the photon theory of light to explain why increasing light intensity increases the number of electrons emitted per second.

Answer 15

→ increasing intensity means more photons interact with more electrons
→ hence more photoelectrons are emitted

Question 16

State the equation that links maximum EK of photoelectrons and photon energy.

Answer 16

hf = work function + KE(max)

Question 17

State the relationship between wavelength and photon energy. State the equation.

Answer 17

→ E inversely proportional to λ

→ E = hc/λ

Question 18

State what is meant by the stopping potential

Answer 18

the minimum potential difference that prevents photoelectrons from crossing the gap between two metal plates

Question 19

State the relationship between potential difference and the force experience by photoelectrons.

Answer 19

greater the potential difference, greater the force of repulsion

Question 20

Describe & explain an experimental set up to stop photoelectric emission.

Answer 20

→ potential difference put across two plates
→ top plate is negative and so repels photoelectrons

Question 21

What is the work done on photoelectron travelling between two plates equal to?

Answer 21

initial kinetic energy of photoelectron

Question 22

State an equation used to calculate the work done on a photoelectron.

Answer 22

work done = electron charge x stopping potential

Question 23

In a vacuum photocell setup, when will the ammeter show a non-zero reading?

Answer 23

when there is a flow of electrons which will happen if:
→ frequency > threshold frequency
→ potential difference < stopping potential

Question 24

State the relationship between stopping potential and frequency. explain why

Answer 24

greater the frequency, greater the stopping potential
→ higher photon energy therefore higher max kinetic energy

Question 25

What did Milikan’s results provide?

Answer 25

→ experimental evidence for the relationship between frequency and stopping potential
→ confirmed Einstein’s photoelectric equation
→ gradient of line can be used to find Planck’s constant

Question 26

Which value can be calculated using Milikan’s results? How?

Answer 26

Planck’s constant, given by the gradient of a freq. vs stopping potential graph

Question 27

State the meaning of the term photoemissive.

Answer 27

a material is photoemissive if it is able to emit electrons due to incident radiation

Question 28

State how to convert from electron volts to joules.

Answer 28

multiply by electron charge

Question 29

What is one electron volt equal to? (in words)

Answer 29

the work done on an electron as it moves through a potential difference of one volt

Question 30

State how Louis de Broglie produced a diffraction pattern to observe electron diffraction.

Answer 30

→ a beam of electrons is directed at a thin graphite film
→ inside a vacuum tube

Question 31

Describe the diffraction pattern observed by Louis de Broglie and what it proved.

Answer 31

→ pattern observed as series of concentric rings
→ similar to diffraction grating pattern
→ showed that e- behaved as waves

Question 32

State and explain the effect of increasing accelerating voltage on the diameter of rings produced in electron diffraction.

Answer 32

→ increasing accelerating voltage increases kinetic energy
→ wavelength decreases
→ diameter decreases

Question 33

Explain why graphite is ideal for electron diffracion.

Answer 33

→ crystalline structure
→ gaps between atoms act as slits allowing e- to spread out

Question 34

Describe the experimental setup to demonstrate electron diffraction.

Answer 34

→ e- are accelerated in an electron gun to a high potential
→ directed through thin film of graphite
→ e- diffract from gaps between carbon atoms
→ produce a circular pattern on a fluorescent screen made from phosphor

Question 35

State the effect of changing momentum, in terms of wavelength, on the amount of diffraction.

Answer 35

→ higher momentum = shorter wavelength
→ small λ means light spreads out less
→ hence a smaller radius is produced