Hsc Physics Topics

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HSC Physics Option Topic

From QUANTA to QUARKS
What is this topic about?
To keep it as simple as possible, (K.I.S.S.) this topic involves the study of:
1. RUTHERFORD & BOHR MODELS OF THE ATOM
2. DE BROGLIE & MATTER WAVES
3. INTO THE NUCLEUS
4. APPLICATIONS OF NUCLEAR PHYSICS
...all in the context of the history, nature and practice of Physics.

1. RUTHERFORD & BOHR MODELS OF THE ATOM
What Has Gone Before...

The Rutherford Model of the Atom

The entire Science of Chemistry and much of
Physics is built on the foundation of Atomic
Theory... the concept that all matter is composed of atoms.

In 1911, Ernest Rutherford carried out an experiment which indicated that the positively charged part
These quantities of energy are “quantised”, so each orbit is really a “quantum energy level” within the atom. (410 nm nanometres)

b) Use the “Wave Equation” to find the frequency. c = λ.f
3.00x108 = 4.10 x10-7x f
∴ f = 3.00x108/4.10x10-7
= 7.32x1014Hz.

The amount of energy absorbed or emitted during a
“jump” is defined by Plank’s Equation E = hf, and the corresponding wavelengths of light are defined by the Rhydberg Equation. The integer numbers nf and ni turn out to be the “quantum numbers” of the orbits, counting outwards from the nucleus.

c) Use Plank’s Equation to calculate the energy carried by one photon of light in the Hδ spectral line.
E = h.f
= 6.63x10-34 x 7.32x1014
= 4.85x10-19 J.

• Electrons in “allowed orbits” have quantised amounts of angular momentum too.
Bohr figured out that the amount of angular momentum possessed by an electron must always be a multiple of h/2π. The significance of this will be dealt with in a later section. HSC Physics Option Topic “From Quanta to Quarks”
Copyright © 2006-2009 keep it simple science
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De Broglie began from Bohr’s equations which showed that (as a particle) the angular momentum of the electron would be a multiple of h/2π.

Diffraction is something that only waves do.
Barrier

From this he was able to show that (when showing its wave properties) the electron would have a wavelength related to its mass and velocity: with gaps in it

Parallel wave fronts approach the barrier. λ= h mv Most of the wave energy will be absorbed or reflected. λ = wavelength (metres) of the electron. h = Plank’s constant (= 6.63x10-34) -31 m = mass of the electron (= 9.11x10 kg) v = velocity of the electron, in ms-1.

The part of the wave which gets through a gap will act like a point source of waves. A semicircular wave pattern forms from each gap.
This is
Diffraction

Example Calculation
Find the wavelength of an electron which is
5
-1 travelling at a velocity of 4.35x10 ms .
Solution

λ= h mv You can see diffraction occur if you watch water waves enter a harbour or similar.

= 6.63x10-34/(9.11x10-31 x 4.35x105)
= 1.67x10-9 m
(1.67 nanometres)

HSC Physics Option Topic “From Quanta to Quarks”
Copyright © 2006-2009 keep it simple