MODULE 7 IQ3

Question 1

what are blackbodies

Answer 1

emits EMR over a wide range of wavelengths
- the dominant wavelengths of an object will largely depends on its surface temperature
- idealistic object that absorbs all EMR that is incident on it –> purely theoretical
- because they absorb all EMR, they absorb all wavelengths –> produces a continuous spectrum

Question 2

what is blackbody radiation

Answer 2

• at a specific temperature, the relative intensities and wavelengths emitted by the blackbody changes
• each temperature has a corresponding peak intensity, and the wavelength at which this occurs is the dominant/peak wavelength
• as we move past the peak wavelength, a longer wavelength corresponds to a lower intensity
• the cooler the star is, the longer the wavelength of the dominant colour is

Question 3

wein’s law theory

Answer 3

the higher the surface temperature of the object, the shorter the dominant wavelength is
- thus by determining a star’s dominant wavelength, we can approximate its surface temperature in Kelvin

Question 4

wein’s law formula

Answer 4

λ = b/T

Question 5

UV catastrophe

Answer 5

known problem of black body radiation
- classical model for BBR is the Rayleigh-Jean’s Law and is based on thermodynamics alongside the classical model of light
- model agrees with experimental data for longer wavelengths of BBR, however there is a fatal discrepancy at shorter lengths –> the model predicts infinite intensity of radiation emission when the peak wavelength is int he ultraviolet range
- violates law of conservation of energy

Question 6

planck’s solution to the UV catastrophe

Answer 6

suggested that blackbodies were cavaties that caused specific frequency atomic oscillations, and that radiation that emitted from these oscillations would have a discrete value of energy
- thus, when blackbody emits or absorbs radiation, it has to be an integer multiple of this quantised value, and if it was not, it would not be absorbed or emitted

Question 7

planck’s solution equation

Answer 7

E = nhf
h = 6.626 x 10^-31 Js

Question 8

continuous spectrum?

Answer 8

a black body with thermal equilibrium will emit EMR in quantised values
- distinction between wavelengths emitted by black bodies are so small that it appears to be a continuous spectrum

Question 9

are stars approximate black bodies

Answer 9

stars are treated as approximate black bodies, since they mainly emit reflected EMR
- because stars have high pressures and temperatures, produce a continuous spectrum

Question 10

what is the photoelectric effect

Answer 10

phenomenon when electrons are released from a metal’s surface when light is incident upon it

Question 11

hertz’s observations on the photoelectric effect

Answer 11

observed that higher frequency EMR would induce stronger sparking
- if he used glass to filter out UV light, the lower frequency EMR could not produce sparking even if he increased its intensity
- when quartz is used to filter out lower-frequency EMR, sparks could be produced, and the higher the intensity of the UV light, the higher the intensity of the sparking
- these observations conflict with maxwell’s wave theory which says that the energy of a wave is related to its intensity or frequency

Question 12

what are cathode ray tubes

Answer 12

• electrons at the cathode, in a cathode ray tube are ‘boiled off’
• these electrons move towards a positive anode, converting electrical energy into KE

Question 13

photoelectric effect in cathode ray tubes

Answer 13

• for each metal surface, electron emission only occurred if the frequency of incident light exceeded a certain threshold
• if photoemission occurred, increasing the intensity of light only increased the number of electrons emitted, but not the KE of the electrons emitted
• increasing frequency past the threshold increased the KE of the electrons emitted
• the stopping voltage is equivalent to the work being done on the electrons by opposing the electric field, alongside the KE of electrons being the work done on them: K(max) = qV(stopping)

Question 14

classical predictions of photoemission

Answer 14

• no threshold frequency as light is dependent on intensity
• if light did not have sufficient intensity/energy there would be a ‘time delay’ until the emission
• if photoemission occurred, higher-intensity light would increase the number electrons emitted (photocurrent)
• if photoemission occurred, higher intensity light would increase the max KE of electrons
• light of any intensity and frequency would be able to release photoelectrons

Question 15

experimental results of photoemission

Answer 15

• there is a threshold frequency for every type of metal
• photoelectric emission was either instantaneous or not observed
• if photoemission occurred, higher-intensity light would increase the number of electrons emitted (photocurrent)
• if photoemission occurred, higher frequency light would increase the max KE of electrons
• only light of a certain frequency would be able to release photoelectrons

Question 16

kinetic energy of electrons

Answer 16

• by applying an electric field in the opposite direction to the electron flow in the cathode ray tube, we can adjust the electric field strength such that at a certain voltage, no electrons are able to reach the anode
• known as stoping voltage and correlates with the maximum KE of electrons emitted

Question 17

einstein’s solution to Planck’s black body radiation

Answer 17

• BBR is emitted in discrete packets of energy with energy, E = hf
• proposed that all EMR was made up of packets of energy known as photons, which each carried energy E = hf
• quantum model –> as energy was no longer dependent on intensity, however it was now proportional to frequency

Question 18

einstein’s experimental observations

Answer 18

• suggested that photons interacted with electrons in a 1:1, all or none interaction
• for photoemission to occur, a photon must have energy that equals or exceeds this binding energy, known as a work function of the metal
• if photon has sufficient energy, it gives all of its energy to 1 corresponding electron, whilst extra energy is used as kinetic energy for the electron
  -if photon doesn’t have enough energy, it does not give any energy at all and emission does not occur

Question 19

einstein’s equation

Answer 19

KE = hf - W
h = 6.626 x 10^-34 Js