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Recap from AS:
Einstein's theory of special 1.When a particle and its
relativity: corresponding antiparticle meet,
E=MC^2 (C=3*10^8ms)
Mass of an object increase
or decreases when it gains
Energy and mass they annihilate each other and 2
gamma photons are produced,
each of energy mc^2 (m is the
or loses energy. mass of particle/antiparticle.
APPLIES TO ALL ENERGY Energy changes in reactions: 2.A single photon of energy in
CHANGES OF ANY Reactions on a sub-nuclear excess of 2mc^2 can produce a
OBJECT particle and antiparticle, each of
level involve significant mass m( pair production)
changes of mass.
Energy released Q=delta mc^2
A DECAY: the nucleus recoils when the a
particle is emitted-so the energy released is When energy is released the
shared between the a particle and nucleus. mass after the change is
­ use conservation of momentum to show always less than the total mass
energy released is shared.
before the change some of the
B DECAY: the energy released is shared mass is changed into energy)
in variable proportions between the B
PARTICLE, NUCLUES,
NEUTRINO/ANTINEUTRINO. If the b
particle has maximum kinetic energy the 1u=1.661*10^-27kg.
neutrino has negligible kinetic energy. The b
particle will always have less kinetic energy E=1.661*10^-27kg * C(3*10^-8ms)
than the energy released because of the
recoil of the nucleus.
=1.49*10^-10J
ELECTRON CAPTURE: nucleus
emits a neutrino which carries away all =931.3MeV
energy released. Also emits a X-ray photon-
when de-excitation.…read more

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NOTES:
STRONG NUCLEAR FORCE-attractive force
When calculating Q in beta decay assume
between any 2 nucleons in the nucleus. neutrino has negligible mass.
Strength?: estimate the force of repulsion If mass of each atom is given instead of mass
between 2 protons at distances of 1fm apart of its nucleus calculate mass of each nucleus
(=10^-15). Between 2 protons= 200 N, so strong by subtracting mass of the electrons from
nuclear must be at least 200 N. mass of each atom.
Range?: no more than 3-4*10^-15m. The
diameter of nucleus can be found by electron
scattering-shows even spacing so strong nuclear
must act between nearest neighbouring
nucleons.
Energy to remove?: in MeV; force is about 200
N, over about 2-3*10^-15 m. SO work
done=7810^-13J.
= 200N * 3.5*106-15M.
=4 MeV.
Repulsion?: become repulsive at 0.5fm or less. If
not the nucleons would pull closer and closer
and the nucleus wouldn't be the same size.…read more

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Binding energy of a nucleus: work that must be done to separate a nucleus into its constituent
neutrons and protons.
To separate a nucleus into
its separate nucleons
work must be done to
overcome the strong
nuclear force. The
BINDING ENERGY
potential energy of
each nucleons Is
When a nucleus forms energy
therefore increased
when removed from is released, as the strong
its nucleus. nuclear force does work pulling
nucleons together, ENERGY
RELEASED =BINDING
ENERGY.
Because energy is released
when nucleus forms, the
nucleuses mass is less than
the mass of its separated
nucleons.
MASS DEFECT: difference between mass of the separated nucleons with the mass of
the nucleus.
MASS DEFECT for =Zmp+(A-Z)mn-Mnuc.
mp-mass of proton, mn-mass of nuetron, Z-protons,A-Z-neutrons, Mnuc-mass of
nucleus.
The mass defect is due to energy released when the nuclues is formed from separate
neutrons and protons.
Binding energy=mc^2…read more

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A particle tunnelling
1. If 2 protons and 2 2. Why?: A particles binding 3. The potential energy of
neutrons bind energy is very large the particle varies with the
together in a big (7MeV per nucleon) distance it is to the outside
enough nucleus, So the particle gains enough the nucleus from inside.
then they emitted kinetic energy to give it
as a A particle enough probability of
quantum tunnelling
4. The gain of kinetic energy when
the A particle forms is not enough
to over come the coulomb barrier
but due to the WAVE NATURE of
A particle it has the probability of
tunnelling through the barrier.
NOTES:
The coulomb barrier is due to the electrostatic
force on the A particle.
The `well' is due to the strong nuclear force.…read more

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Nuclear stability
Binding energy PER NUCELON:
average work done per nucleon
= MORE BINDING
to remove all nucleons from a
ENERGY
nucleus=MEASURE OF
REQUIRED=
STABILITY
MORE STABLE.
FUSION?: process of making
small nuclei fuse together to form
a larger nucleus.
Product has more energy than
smaller nuclei. ­so binding energy
of each nucleon INCREASES
FISSION?: process in which a
large unstable nucleus splits into 2
fragments which are more stable-
Binding energy per nucleon
INCREASES.…read more

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Comments

Uman Nulla

Didn't like the rushed formatting. Good content for 17 slides though

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