Nuclear physics - Radioactivity Flashcards
Describe setup of Rutherford’s α scattering experiment, inc explanation for each point (esp explaining details abt α particle). 5
Narrow beam of α particles (bc +ve), all of the same Ek (bc otherwise slow α particles would be deflected more than faster α particles on same initial path), in a vacuum (bc otherwise α stopped by air particles), fired at a thin gold foil (bc otherwise α scattered more than once). Detectors can move around and detect scattered α particles. α source must have long half life otherwise readings would be lower than earlier readings due to radioactive decay.
Results of Rutherford’s experiment? 2
-most α passed straights through foil with little/no deflection (about 1/2000 deflected).
-small percentage of α (1/10,000) were deflected through angles > 90°.
Interpretation of results of Rutherford’s experiment? 2
-most of atom’s MASS is concentrated in a small region, the nucleus, at the centre of the atom
-nucleus is +vely charged to repel α particles (+ve) that approach too closely.
Paths of some α particles which pass near a fixed nucleus.
Draw:
A
B
C
D
E
A, B, D deflected at diff angles. the closer the α particle, the greater its deflection bc electrostatic force of repulsion increases with decreasing distance between them. (A barely deflected, B deflected backwards, D deflected forwards).
C collides head-on w the nucleus and rebounds (deflected at 180°).
E doesn’t approach nucleus closely enough to be deflected.
Estimating size of nucleus:
1: 1/10000 α particles deflected by more than 90°.
2: Thin foil so each α scattered once.
3: A typical value for number of layers of atoms is 10^4
The probability of an α particle being deflected by a given atom is 1/10000n, where n is the number of layers of atoms. This probability depends on the effective cross-sectional area of the nucleus to that of the atom.
For a nucleus of diameter d in an atom of diameter D, the area ratio si equal to 1/4πd^2 / 1/4πD^2 = d^2/D^2
∴ d^2 = D^2/10000n.
A typical value for n = 10^4 gives d = D/10000
Rutherford found that radiation… (3 properties - found via experiment).
-ionises air, making it conduct electricity .
-was of two types: α more easily absorbed, β more penetrating (γ discovered a year later).
-magnetic field deflects α and β in opposite directions, and has no effect on γ - α is +ve and β is -ve (γ later shown to consist of high energy photons).
Radioactivity experiments: 4
Ionisation
Cloud chamber observations
Absorption tests
Range in air
Ionisation - explain (clue pA)
Ions created are attracted to oppo charge electrode where they’re discharged. E-s pass through pA as a result of ionisation. I ∝ number of ions created per second ∴ can see:
-α most ionising and if move source a few cms away the current ceases,
-β weaker ionisation and range varies up to a metre or more,
-γ least ionising bc photons carry no charge
Cloud chamber observations
(α vs β tracks and explain)
Contains air saturated w a vapour at a very low temp. Due to ionisation of air α/β passing through leave visible track.
α: straight track, easily visible, same length track if same isotope (α have same range). α particles from given isotope always emitted w same Ek bc α particles and nucleus move apart w equal and opposite momentum.
β: wispy tracks, easily deflected when collide w air molecules, less ionising ∴ harder to see tracks. in β decay electron antineutrino emitted as well ∴ the nucleus, β and neutrino share the energy released in variable proportions.
Absorption tests
Using Geiger tube and counter to find corrected count rate of source.
Keeping distance constant, can measure absorbance by using no absorber then absorber of diff thickness etc
Range in air - how and explain observations
Geiger tube and counter- vary distance.
-α few cm then sharp decrease in count rate bc all same Ek
-β up to a metre, gradual decrease bc range of Ek up to a max
-γ unlimited range in air, gradual decrease in count rate bc radiation spreads out in all directions. Proportion of γ photons from source entering tube decreases according to inverse square law (all same E).
β radiation consists of fast moving e-s. How was this proven?
By working out specific charge of β using electric and magnetic fields - same as e- specific charge.
γ radiation consists of photons with wavelength of… ?
How was this discovered?
10^-11 or less.
Discovery made by using a crystal to diffract a beam of γ radiation.
Inverse square law for γ radiation:
Intensity of radiation =
-Intensity, I, of radiation = radiation energy per second passing normally through unit area.
-For a point source emitting nγ photons per sec, radiation energy per sec = nhf
-At a distance r from source, all photons emitted pass through total area 4πr^2
∴ I=nhf/4πr^2 =k/r^2 ∴ I ∝1/r^2
why might results of experiment not follow expected inverse-square law
-random nature of radiation count
-dead-time in G-M detector
-d is not the real distance between the source and detector
-assume no absorption between source and detector
