Module 4 Photoelectric effect

Question 1

Describe the gold-leaf electroscope

Answer 1

a metal plate (can be made of any kind of metal) attached to a brass rod with a gold leaf which is free to move (attached to the rod )

Question 2

Describe how the photoelectric effect can be demonstrated

Answer 2

using an apparatus called the gold-leaf electroscope - add a charge to the plate on the top (can be done using rabbit fur or sending a current directly through the apparatus)

The brass rod and gold leaf become negatively charged. The gold leaf has a very low mass (very thin) and like charges repel so the leaf rises

Weak UV light is shone onto the plate on top and the leaf falls quickly

The UV light provides energy for the electrons to escape

A powerful visible light source does not release electrons so the gold leaf does NOT fall

Question 3

What part of the gold-leaf experiment cannot be explained using wave theory?

Answer 3

1. There is a threshold frequency below which no electrons are emitted. According to wave theory an intense white light carries more energy then weak UV light but only UV light emits electrons/displaces them
2. The max KE of the electrons does not change when increasing intensity of source. KE of electrons is not dependent on source intensity .It should provide more energy for electrons to escape with.

Question 4

What is a photon?

Answer 4

a quantum of energy of electromagnetic radiation (can be described as discrete energy or as a packet of energy)

Question 5

What does Einstein’s photon model suggest about electromagnetic waves?

Answer 5

EM waves travel in photons which are each a quantum of energy

and that the energy of a photon is directly proportional to it’s frequency E=hf, where h is the Planck constant

Question 6

Describe the properties of photons

Answer 6

Have zero rest mass and charge

Travel at c in a vacuum

Are not effected by electric or magnetic fields

Question 7

What is E=hf calculating or E=hc/λ ?

Answer 7

Energy of a single photon

Question 8

Explain the observations in a gold leaf electroscope using the photoelectric effect

Answer 8

A one-to-one interaction between photons and free electrons of the surface of a metal plate on the top of the electroscope

A single electron absorbs a single photon and gains the energy of a photon according to E=hf

The electron needs a minimum energy to escape, this is called the work function of the metal - Φ

Any remaining energy is the KE of the electron

Electrons are lost so less repulsion between gold leaf and brass rod so gold leaf on electroscope falls

If photon energy hf is less than the work function then NO electrons are emitted

Question 9

What is the work function of a metal?

Answer 9

the minimum amount of energy required for an electron to be removed from a metal/escape

Question 10

Give the word equation of the photoelectric effect observed using a gold leaf electroscope

Answer 10

energy of a single photon = minimum energy required for electron to escape + KEmax of emitted electron

Question 11

What does the work function depend on?

Answer 11

the type of metal

Question 12

What is required for electrons to be released from the metal (PE effect)?

Answer 12

for the energy of a photon to be ≥ the work function

Question 13

What is the threshold frequency?

Answer 13

the minimum frequency of EM radiation that will remove electrons from the surface of the metal plate

Question 14

How is the threshold frequency written?

Answer 14

fo

Question 15

What is the mathematical relationship between the threshold frequency and the work function?

Answer 15

hfo = work function

Question 16

What happens to the electrons when photons are at the threshold frequency? (PE effect)

Answer 16

they escape with zero KE

Question 17

What is the threshold wavelength?

Answer 17

the longest possible wavelength that will release electrons (E=hc/λ)

Question 18

Why is the KE of removed electrons described as a range? (PE effect)

Answer 18

Some electrons may lose KE due to collisions as they escape the metal

The electrons with maximum KE are emitted at the very surface of the metal so no collisions and escape with all their KE

Question 19

What is the range of KE of emitted electrons? (PE effect)

Answer 19

KE = 0 to KEmax

Question 20

Explain how an electron may escape with zero KE? (PE effect)

Answer 20

Photon is absorbed by an electron which is located inside the metal

Electron gains energy in the form of KE, but then loses this energy due to collisions

By the time it reaches the surface of the metal all KE is lost due to collisions and electrons just escape

Question 21

What electrons will always have a KE<KEmax? (PE effect)

Answer 21

electrons which are not on the surface of the metal

Question 22

Describe the relationship between intensity and the number of photons emitted in a beam of EM radiation per second (PE effect)

Answer 22

I=P/A (when calculating area remember to square as area is in m^2)

Power is rate of work done per second, work done is in J - so P=energy transferred per second

P=hfxN - where N is number of photons emitted per second and hf is energy of one photon

Therefore I is proportional to N

Question 23

What is the effect of doubling the intensity of the radiation landing on a metal plate (PE effect)?

Answer 23

No change to max KE of emitted electrons (photon energy has not changed)

Doubling intensity means number of photons per second doubles

One to one interaction so if number of photons per second doubles then the number of emitted electrons per second also doubles and the photoelectric current doubles

Question 24

What is the effect of increasing the frequency of the radiation incident on the metal plate (PE effect)?

Answer 24

Photon energy increases (E=hf)

If photon energy > work function then electrons emitted with higher maximum kinetic energy.

Work function unchanged.