i just need this physics work done.

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    Introductory Physics Assignment 2
    This test covers material from the modules: ATNU, MATT and HTTH.
    INSTRUCTIONS:
1. Write your name on this paper
2. Sign the declaration
3. Show all working in the space provided    Total Marks = 60
Student Number:      Student Name:     
Introductory Physics
Assignment 2 – Version 7
CQUniversity Australia
10
ATNU – Atomic and Nuclear    20 marks
1. Describe the structure of an atom, the quantised electron shell model and how it produces unique absorption and emission spectra. (Remember to explain ground and excited states, and how and what happens when electrons jump between)
According to bohr’s model, electrons could only orbit the nucleus in specific orbits or shells with a fixed radius. Only shells with a radius given by the equation below would be allowed, and the electron could not exist in between these shells.
By keeping the electrons in circular, quantized orbits around the positively-charged nucleus,
He define the absorption and emission process in terms of structure. Electron absorbs energy in the form of photons to reach high energy level as long as photon’s energy level become equals to the difference between the initial and final energy level. The excited state will be in less stable state
The ex
(4)
2. Given E0
 54.4 eV for a 4 He (an ion with two protons, two neutrons, and one electron),
(4.5)
determine,2
(i) the energy of a photon that will cause the excitation of an electron from level n  1 to
n  5
(ii) frequency associated with this photon
(iii) and wavelength associated with this photon in nanometers
Given
Energy at level 1, E1=  54.4 eV
Solution:
i. Energy at level 5, E5 = E1/(5*5) = -54.4/25 = -2.17 eV
Energy of a photon that will cause the excitation of an electron from level n  1 to n  5,dE = E5 -E1
dE = -2.176 + 54.4 = 51.924 eV
ii. dE = hv
v = Frequency associated with phon
h = 4.14 x 10-15 eV s
v = dE/h
v = 51.924/4.14 x 10-15 = 12.37 x 1015 s-1
iii. Speed of the light, c = 3 x 108 m/s
Wavelenght, lambda = ?
v = c/lambda
lamda = c/v
lambda = 12.37 x 1015/3 x 108 = 4.12 x 107 m
3. a)    The decay series for plutonium-242 to uranium-234 is shown below    (2)
242 Pu stage1 238 U stage2  234 Th stage3  234 Pa stage4  234 U94    92    90    91    92
Determine the type of decay in the four stages shown, by showing the chemical equations at each step
Stage 1: alpha decay
PU24294 = PU23892 + H42 + Q
Stage 2: alpha decay
U23892 = Th23490 + H42 + Q
Stage 3: Electron beta decay
Th23490 = Pa23491 + e0-1 + Q +Ve
Stage 4: Electron beta decay
Pa23491 = U23492 + e0-1 + Q +Ve
b)    List a type of material (for each) that will stop each of the three types of radiation.    (1.5)
alpha radiation can be stop by few cm of air, apiece of paper
Beta radiation can be stop by sheet of plastic or aluminium
Gamma radiation can be stop by led, wall
4. a)    A fossilised tree was tested and contains 10 grams of Carbon-14. Given that there was 12 grams of Carbon-14 present when it died, determine the age of the fossil? (Hint – the half-life of Carbon-14 is stated in the module, from this you can find the decay constant)
(3)
b)    How much Carbon-14 will be present in thesample in 100 years from now?    (1)
Solution:
(a)
half life of the carbon =...