Breakdown of Classical Physics

Question 1

What is Stefan-Boltzmann Law?

Answer 1

Pem = eσAT^4, where e is the emissivity between 0 and 1

Question 2

What is the understanding of a blackbody?

Answer 2

The emissivity is a dimensionless number which has a value between 0 and 1 depending on the specific nature of the surface of the object. A perfect emitter – a so-called a blackbody or backbody radiator – has an emissivity of e = 1.

Thus, a perfect absorber would appear black since it does not reflect any light at all. It is for this reason that these idealised bodies are called blackbodies.

Question 3

What is Wien’s displacement law?

Answer 3

λmaxT = 2.898 × 10^−3 m K -> the total power being emitted was given by the Stefan-Boltzmann law equation at a given temperature T (in kelvin) the wavelength λmax (in metres) at which the maximum intensity emissions are observed is given by the relationship

Question 4

What is Planck’s law and the assumptions?

Answer 4

I(λ, T) = (2πhc^2) / (λ^5(e^(hc/(λkT))− 1)) -> he assumed that the blackbody radiation came from atoms within the walls of the cavity
oscillating back an fourth – so-called oscillators
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Assumption 1: The energy of an oscillator can only have certain discrete values En: En = nhf
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Assumption 2: An oscillator can only emit or absorb energy in units of
E = hf

Question 5

What is the understanding of photons, its formula and its properties?

Answer 5

That the radiation itself consists of particles only – now known as photons – and that light with frequency f is just made up of many discrete particles each with energy -> E = hf

Light was incident on certain metallic surfaces. electrons would be emitted from those surfaces. This is known as the photoelectric effect and the ejected electrons are referred to as photoelectrons.

These experiments showed that the photoelectrons have the following key properties:

1. The kinetic energy K of the photoelectrons have a range of values, but it is always less than some maximum value Kmax (which depends on the particular metal).
2. The maximum kinetic energy of the ejected electrons Kmax is independent of light intensity. The intensity is proportional to the number of electrons emitted from the surface.
3. ## The maximum kinetic energy Kmax of the ejected electrons is proportional to the frequency f of the electromagnetic radiation, and below some cutoff frequency fc no electrons are ejected no matter how intense the light.The maximum kinetic energy Kmax of the ejected electrons occurs for the most loosely bound of the electrons. We define Φ as the binding energy of the most weakly bound electrons – also known as the metal’s work function – and so
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   Kmax = h f − Φ . Maximum kinetic energy of photoelectrons

Question 6

More photons in relation to the Compton Effect

Answer 6

E = hf = hc /λ ->Energy of a photon

p = E/c -> Momentum of a photon

p =h/λ -> Momentum of a photon in terms of wavelength

When this experiment is carried out at various angles θ, x-rays arrive at the detector with two distinct wavelengths:
(i) a wavelength λ0 – the same wavelength as the original incident radiation;
(ii) a wavelength λ0 which is longer than the original incident radiation, i.e λ0 > λ0.

• Photon before collision: The energy of the photon is E0 = hf0 = hc/λ0 and the magnitude of its momentum is p0 = h/λ0
• Electron before collision: The electron is at rest and so has zero kinetic energy and zero momentum.
• Photon after collision: The photon is scattered at an angle θ and has energy E0 = hf0 = hc/λ0 and momentum with magnitude p’ = h/λ0
• Electron after collision: The electron recoils, carrying away some energy and momentum. It is not detected in the experiment.

Question 7

What is the Compton shift equation

Answer 7

λ’ − λ0 = (h / (me c)) (1 − cos θ)

Question 8

Bohr’s Hypothesis

Answer 8

An electron of mass me moving in a circle of radius r at speed v about the nucleus has an angular momentum of magnitude L about the nucleus given by: L = mevr

So Bohr’s hypothesis that L must come in units of ~ is the statement that mevr = n~ n = 1, 2, 3, . . . .

vn, rn and En -> you should be able to show that the allowed speeds vn, the radii of orbits rn and energy En of the electron are quantised (REFER TO NOTES)

En = -13.606eV (Z^2 / n^2)

Question 9

What is de Broglie wavelength

Answer 9

that matter particles with a momentum of magnitude p could have an associated wavelength λ given by -> λ =h/p

Question 10

Bohr’s Model

Answer 10

a single electron of mass me moving in a circle of radius r at speed v about the nucleus has an angular momentum of magnitude L about the nucleus which comes in units of ~: L = mevr = n~ n = 1, 2, 3, . . . .