Module 4 Photoelectric effect COPY Flashcards

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1
Q

Describe the gold-leaf electroscope

A

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 )

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2
Q

Describe how the photoelectric effect can be demonstrated

A

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

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3
Q

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

A
  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.
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4
Q

What is a photon?

A

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

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5
Q

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

A

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

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6
Q

Describe the properties of photons

A

Have zero rest mass and charge

Travel at c in a vacuum

Are not effected by electric or magnetic fields

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7
Q

What is E=hf calculating or E=hc/λ ?

A

Energy of a single photon

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8
Q

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

A

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

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9
Q

What is the work function of a metal?

A

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

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10
Q

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

A

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

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11
Q

What does the work function depend on?

A

the type of metal

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12
Q

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

A

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

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13
Q

What is the threshold frequency?

A

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

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14
Q

How is the threshold frequency written?

A

fo

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15
Q

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

A

hfo = work function

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16
Q

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

A

they escape with zero KE

17
Q

What is the threshold wavelength?

A

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

18
Q

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

A

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

19
Q

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

A

KE = 0 to KEmax

20
Q

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

A

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

21
Q

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

A

electrons which are not on the surface of the metal

22
Q

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

A

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

23
Q

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

A

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

24
Q

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

A

Photon energy increases (E=hf)

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

Work function unchanged.

25
Q

Photoelectric effect - graph of maximum kinetic energy against frequency
Shape?
Main features?

A

Straight line y=mx+c
KEmax= hf - work function
y intercept is minus work function
x intercept is threshold frequency
Gradient is Planck constant h