13.2.3 - 13.2.4

13.2.3
Describe one piece of evidence for the existence of nuclear energy levels:


The energies of both alpha and gamma are discrete and have only certain values. Since these discrete energies come from the nucleus, it must mean that the nucleus has discrete nuclear energy levels (just like electrons do).

13.2.4
Describe β+ (positron) decay, including the existence of the neutrino:


When energy calculations were done with radioactive beta emissions, it was observed that the beta particles had lesser kinetic energy than they would have expected (after using more detailed levels of calculations). This opposed the law of conservation of energy as the energy had to come from somewhere. In 1930, Wolfgang Pauli said that there was a very small particle that therefore carried away this missing kinetic energy and momentum. This particle was named the neutrino. 

13.2.1 - 13.2.2

13.2.1:
Explain how the radii of nuclei may be estimated from charged particle scattering experiments:


In charged particle scattering experiments, the particle that is repelled straight back will be at rest for a very short amount of time before it changes direction and travels back to where it came from. Rutherford proposed that this piece of information can be used to calculate and find the approximate size of the nucleus. This is because at that point in time where the particle is at rest with a velocity of 0, its kinetic energy will be exactly balanced out by its 
electrical potential energy due to the electrostatic force which repels the nuclei.


13.2.2: 
Describe how the masses of nuclei may be determined using a Bainbridge mass spectrometer:

A magnetic field is used to deflect the ions of a substance. If the ions have the same charge they will have the same velocity. The radius of their circular path however depends on how heavy that particular ion is. A larger mass ion will travel a larger distance (larger circle radius). Therefore, the masses of nuclei may be determined using a Bainbridge mass spectrometer.

Nelson Questions

19.4
Max Kinetic Energy = hf - ɸ
= (6.63x10^-34 x 3.19x10^-9) - 3.413x10^-19
= -3.8x10^-19
Since Kinetic energy = Vsq
3.8x10^-19/9.1x10^-31 = 4.176x10^11  eV


19.5
Max Kinetic Energy = hf - ɸ
2.4x10^-19 = 6.63x10^-34 x (3x10^8/500x10^-9) - 
ɸ = 1.56x10^-19

12.2.1 to 12.2.5

12.2.1
 Describe the emf induced in a coil rotating within a uniform magnetic field

12.2.2 

Explain the operation of a basic alternating current (ac) generator

• A coil of wire rotates in the magnetic field due to an external force. As it rotates, the flux linkage of the coil changes with time and induces and e.m.f (refer to Faraday's law) causing current to flow.
  • Sides the 2 sides parallel with the magnet will experience a force opposing the motion. (refer to Lenz's Law)
  • The work done rotating the coil generates electrical energy.

12.2.3

Describe the effect on the induced emf of changing the generator frequency

• The speed of rotation is changed it will affect:
  Time (distance between peaks)
  EMF (amplitude of peaks)
• For example, if the speed is doubled therefore, the time period will be halved and the EMF (amplitude) will be doubled.

12.2.4

Discuss what is meant by the root mean squared (rms) value of an alternating current or voltage

• When the output of an a.c. generator is connected to a resistor, an alternating current will flow. A sinusoidal potential difference means a sinusoidal current. 

12.1.1 to 12.1.4

12.1.1 Describe the inducing of an emf by relative motion between a conductor and a magnetic field: All conductors contain free electrons. As the free electrons move downwards through the magnetic field it will experience a force. Using Flemming's left had rule we can deduce the direction of the force. This force will cause the free electrons move in the direction of the force. Since the electrons are moving towards a particular side of the conductor, this has caused the lattice atoms on the opposite side to the force to become positive and therefore there is a potential difference between the 2 ends of the conductor. 12.1.2 Derive the formula for the emf induced in a straight conductor moving in a magnetic field: Fb = Fe Fb = Bev E = - dV/dx potential gradient = V / L Fe = Ee = Ve / L Ve / L = Bev V = BLv Therefore emf induced = BLv 12.1.3 Define Magnetic Flux and Magnetic Flux Linkage: Magnetic flux is the strength of a magnetic field over a given area A flux passing through a coil of N turns of wire is said to have a link with the coil.Therefore Magnetic Flux Linkage is given by: Flux Linkage = Number of coil turn x Magnetic Flux 12.1.4 Describe the production of an induced emf by a time-changing magnetic flux: ·         An e.m.f. is also produced in a wire if the magnetic field changes with time. The universal rule that applies  to all situations involving emf is stated as: "the magnitude of an induced e.m.f is proportional to the rate of change of flux linkage" State Faraday's law and Lenz's law: ·         Faraday's law:   ·         Lenz's law states that "The direction of the induced e.m.f. is such that if an induced current were able to flow, it would oppose the change which caused it".   Questions Q39 a) emf = BLv               =50x10^-6 x 0.2 x 20 = 2 x 10^-45 b) I = V / R             I=2x10^-4 / 2        = 1x10^-4 c) I^2 R = 1 x 10^-4 x 2 = 2 x 10^-8 W d) Work done = energy dissipated = 2 X 10^-8 e) s = d/t s = 20/1 = 20m/s f) F = work / d = 2x10^-8/20 = 1x10^-9 Q40 a) Total flux = 5000 x 10^-6 x 2 x 10^-4 = 1 x 10^-6 2 x 10^-4 m^2 = 2 cm^2 b) would be a quarter since flux density would be halved and time would be doubled = 0.25 X 10^-6tm^-2s^-1 c) since induced emf = rate of change of flux, the induced emf is 0.25 x 10^-6 V Q41 a) = 6 x 10^-4 x 50 x 500 x 10^-6 therefore total flux = 1.5 x 10^-5 tm^-2 b) Φ = BcosθxA Φ = 500 x 10^-6 x cos30 x 6x10^-4 x 50 Φ = 1.3 x 10^-5 tm^-2 c) Blv = tm^2/s 1.3 x 10^-5 / 3 = 4.3 x 10^-6 V