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1.1-Inside the atom AQA AS Physics P1 Matter And Radiation Kerboodle Answers
1.Ans-
1 a) i) 6,6 ii)8,8 iii)92, 143 iv)11,13 v)29,34
b) i) Uranium ii)Oxygen and carbon.
2.Ans-
Neutron
Electron
Proton
A 63Cu29 atom loses two electrons.For the ion formed
Calculate its charge in C
State the number of nucleons it contains.
Calculate its specific charge in Ckg-1
Ans-
43.2X10-19 C
63
4.1X107 C/kg
3.Ans-
2.66X10-26
O (16) is the most abundant…8 protons, 8 neutrons, 8 electrons.
O (17)- 8 protons, 9 neutrons, 8 electrons.
O(18)- 8 protons, 10 neutrons, 8 electrons.
1.2-Stable and unstable nuclei AQA AS Physics P1 Matter And Radiation Kerboodle Answers
1.Ans-
electrostatic force
strong nuclear force
strong nuclear force
electrostatic force
2.Ans-
Src- https://scontent-bom1-1.xx.fbcdn.net/v/t1.15752-9/41698828_1992355961063790_3276453528746328064_n.jpg?_nc_cat=0&oh=d7d01b25bf3513f07224f124cc7139f5&oe=5C339FDD
.
4.Ans-
The term hypothesis refers to an idea that is suggested as the possible explanation for something but has not yet been found to be true or correct.
i) They are electrically neutral and interact only via the weak interaction.
ii) Uranium and Thorium or Sun and Supernova
1.3-Photons AQA AS Physics P1 Matter And Radiation Kerboodle Answers
1.Ans- a) Gamma rays,Xrays,UV rays,Visible light,Infrared,Microwave,Radio
b)
Src- https://scontent-bom1-1.xx.fbcdn.net/v/t1.15752-9/41637636_1968338159915523_5078757123387555840_n.jpg?_nc_cat=0&oh=62c968d990e3822e6a5da8fe2a4c1f98&oe=5C29E32F
2.Ans-
Src- http://hyperphysics.phy-astr.gsu.edu/hbase/electric/imgel2/emwavec.gif
3.Ans-
Src- https://scontent-bom1-1.xx.fbcdn.net/v/t1.15752-9/41637636_1968338159915523_5078757123387555840_n.jpg?_nc_cat=0&oh=62c968d990e3822e6a5da8fe2a4c1f98&oe=5C29E32F
4.Ans-
1.4-Particles and antiparticles AQA AS Physics P1 Matter And Radiation Kerboodle Answers
1.Ans-
936.85Mev
The mass of one proton is 938 MeV/c^2, the anti-proton has the same mass as a proton. So an proton-antiproton pair has a total mass of 1876 MeV/c^2.
Due to the equivalence of energy and mass from the equation E = mc^2, if m is 1876 MeV/c^2, then:
E = 1876 MeV.
So the energy of the photon is 1876 MeV (this is due to conservation of energy)
2Ans- Because for a photon to produce a proton-antiproton pair the minimum energy required is 1876MeV as the mass of one proton is 938 MeV/c^2, the anti-proton has the same mass as a proton.So an proton-antiproton pair has a total mass of 1876 MeV/c^2.Due to the equivalence of energy and mass from the equation E = mc^2, if m is 1876 MeV/c^2, then: E = 1876 MeV.
3.Ans-
a) Energy equivalent of one electron mass is 0.511 MeV. A positron has the same mass as the electron, therefore in your annihilation, you have a total of 1.022 MeV. Assuming linear momentum of the electron and positron is small, then momentum before = 0, so momentum after = 0 too. This is achieved by generating two photons with equal and opposite momenta. So you have two 0.511 MeV photons emitted in diametrically opposite directions
b) i Total energy of the system is 0.511 + 0.511 + 0.158 = 1.180 MeV (mass energy of two electron-like particles + KE of one of them)
ii) Split the total energy between two photons. Energy = 1.18/2 = 0.59 MeV
4.Ans-
(a) A nucleus with too many protons (proton-rich) will be unstable, so a proton will turn into a neutron and emit a positron as a result. This is called beta + decay (opposite of normal beta decay).
(b) 1: Pair production: positron created by two photons colliding. Positron emission: positron created by beta + decay. 2: Pair production: An electron is also created. Positron emission: no electron is created (no other products)
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1.5-Particle interaction AQA AS Physics P1 Matter And Radiation Kerboodle Answers
1.Ans-
Src- https://www.sciwebhop.net/sci_web/physics/a-level/as_module1/images/past_m7.jpg
b)
c)
Src- https://upload.wikimedia.org/wikipedia/commons/thumb/c/ca/Feynmandiagram.svg/1200px-Feynmandiagram.svg.png
2.Ans-
3.Ans-
a)
i)
ii)
b) 0.001fm
speed=dist/time
time=3.33X10-27s
4.Ans-
W-bosons are charged and the photons are uncharged, W-bosons have a non zero rest mass and photons hsve zero rest mass. Also the W-boson is the “exchange particle” in interactions involving the weak nuclear force, but photons are the exchange particle of the electromagentic force.
In 4b a proton changes into a neutron due to weak nuclear forces by emitting a W-boson while in fig 5 it changes by capturing one of the inner shell electron and it is known as electron capture.
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Practice questions-
1Ans-
a)
Src- https://scontent-bom1-1.xx.fbcdn.net/v/t1.15752-9/41660026_704685219894156_1862662058542104576_n.jpg?_nc_cat=0&oh=466d7169233488c890edc7c764d4d60a&oe=5C39B6CD
b)
i) This is because they have opposite charges one has a negative charge while the other has a positive charge.
ii) They will have same amount of kinetic energy assuming that the collision is perfectly elastic.
iii) He deduced so because the particles turned in the opposite directions so he thought they must be having opposite nature of charges.
2.Ans-
90 protons,90 electrons,139 neutrons
Z is 90,X=90,Y can be 225 or 229
3.Ans-
18 protons and 19 neutrons
-3.2X10-19C
C) Ans-
neutron
electron
d) Ans- %=16X9.1X10-31/37X1.6X10-27= 0.91%
4.Ans-
protons=14,neutrons=14,electrons=14(most stable config)
or protons=13,neutrons=15,electrons=13
i) Isotopes are the atoms which have same atomic number but differ in mass number i.e they have different number of neutrons.
ii) 82
iii) 3.89X107
iv) 95
5.Ans-
X=229-4=225
Y=90-2=88
b) 56.25
6Ans-
a)
i) Alpha particles are helium nuclei, containing two positively charged protons and two neutrally/zero charged neutrons, and therefore have an overall charge of 2+.
Properties-
Alpha particles have the least penetration power
Alpha particles have the highest ionization energy
It has a positive charge of 2 units
ii) 218At85 214Bi83 + 4He2
b)
i) Its atomic number increases by 1 and mass number of remains unchanged.
ii) 99Mo42 99Tc43 + Beta- + neutrino
7.Ans-
I)
For this I used c/wavelength = f
3.00×10^8/8.30×10^-13 = 3.61×10^20 Hz
E=hf
E = 6.63×10^-34 * 3.61×10^20 = 2.40×10^-13J
m=E/c^2
m=2.40×10^-13/9.00×10^6 = 2.7×10^-30Kg
ii) 2.40×10^-13J
iii) 1.498MeV
b) weak nuclear forces
8.Ans-
i) electron
ii) annihilation occurs when positron meets its anti particle i.e electron
b) Assuming both the particle and the antiparticle have negligible kinetic energy when they meet, the total energy in the system is 2*(0.51 MeV) = 1.02 MeV. (Antiparticles have the same rest mass as their corresponding “ordinary” particles.)
So the energy released in the annihilation is: 1.02 MeV
9Ans-
a)
i) They are annihilated (1 mark) (OR The mass is converted to electromagnetic energy in the form of a pair of gamma rays (or photons)) The gamma rays will be virtual if the then produce other particles – but may just be gamma rays that we could detect… they are then not virtual. Only call them virtual when they are exchange particles.
ii) On annihilation the rest mass of the particles is converted to energy therefore the total energy after the collision is the KE before plus the mass conversion energy
b) There may be more particles – more massive particles – or particles with a greater kinetic energy.
10.Ans-
Gravitons.Gravitational force in involved when it is produced.
1down vote
An exchange particle plays a very important role in an interaction.It is a virtual particle that mediates the interaction between two other particles. It is virtual because it need not be on the mass shell and hence is not directly observable.
12.Ans- Pair production is a process in which a gamma ray of sufficient energy is converted into an electron and a positron.Examples include creating an electron and a positron, a muon and an antimuon, or a proton and an antiproton.
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References:
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