Solutions for Radiation Safety in Radiation Oncology, 1st Edition Rajan (All Chapters included)
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Course
Radiation
Institution
Radiation
Complete Solutions Manual for Radiation Safety in Radiation Oncology, 1st Edition by K. N. Govinda Rajan ; ISBN13: 9781498762243.(Full Chapters included Chapter 1 to 12)....Basic Atomic & Nuclear Physics. Basic Radiation Physics. Radiation Quantities & Units. Biological effects of Radiation. Radiat...
SOLUTIONS MANUAL FOR
RADIATION SAFETY
IN RADIATION
ONCOLOGY
by
K. N. Govinda Rajan
Complete Chapter Solutions Manual
are included (Ch 1 to 18)
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** All Chapters included
, Chapter 1
Answers
1. For n = 4, l assumes values of 0,1, 2 and 3.
2. M shell
3. ml values identify the orbitals. ml values = - l to + l = (2l+1) = 5values.So,there are 5 orbitals.
4. Only other quantum number remaining is the spin quantum number ms. So, there can be two
electrons with parallel and ant parallel spins.
5. Whichever is closer to the nucleus (n=2 electron).
6. l = 0,1,2,3 i.e. 4 subshells. (no. of subshells = n).
7. Carbon atom has 6 electrons. So, its configuration would be 1s22s22p2.
8. The number of electrons that the sub-levels d, f and g:
Orbitals in sub-shell are determined from l.
l 0 1 2 3 4
Sublevels s p d f g
Orbitals 1 3 5 7 9 (2l+1) orbitals per sub-shell
No. of electrons 2 6 10 14 18
9. The formula 2n2 gives the maximum number of electrons that can occupy the shell of
principal quantum number n. So, for n = 2, 3, and 4, the maximum number of electrons will
be 8, 18 and 32 respectively.
10. RU = 1.2 x 10-15 x (238)1/3 = 7.4366 x 10-15 m.
11. The binding energy per nucleon for a nucleus is determined as follows:
Mass defect = ΔMC = MC - (3 mn + 3 mp)
BE = ΔMC c2
BE per nucleon = BE / mass number of the nuclide
Note:
Mass defect is the mass deficit in the nucleus or the reduction in mass of nucleons during
the nucleus formation and so is a negative quantity. So is the binding energy released
, (exothermic reaction). However, for getting the numerical value, the order of the terms does
not matter.
12. Protons and neutrons are the basic constituents of the nucleus and electrons are found
outside the nucleus.
13. Element: iodine; atomic number = 53; mass number = 125;
14. Element: Fluorine; no. of protons = no. of electrons = 9; no. of neutrons = 19-9 = 10;
15. Pauli’s exclusion principle that applies to all off integral particles (fermions).
16. 1 eV = 1.6 x 10-19 J. By definition, 1 eV is the energy gained by an electron under an
accelerating voltage of 1 volt. i.e. Q x V joules = 1.6 x 10-19 x 1 Joules.
17. We will first express 1 amu in kg.
One gram atom of any element contains 6.023 x 1023 atoms (Avogadro number).
i.e. 6.023 x 1023 atoms weigh 12 g. Let m’ be the mass of one carbon atom in g.
6.023 x 1023 x m’ = 12 g ; m’ = (.023 x 1023) g
By definition, 1 amu = (1/12) x mass of a carbon atom
= (1/12) m’ = (1/ 6.023 x 1023) g = 1.66 x 10-27 kg
E = m’ C2 Joules = 1.66 x 10-27 x (3 x 108)2 J = 14.9 x 10-11
1 MeV = 1.6 x 10-13 J ; i.e {(14.9 x 10-11) / (1.6 x 10-13) }= 931 MeV .
One generally uses a figure of 931.1 MeV.
18. 0. There is nothing to bind. There is only one proton.
19. Part of the nuclear mass is converted to binding energy.
20. The answer to the last question gives us the clue to answer this question. Total mass of all the
nucleons in the atom minus the mass of the nucleus gives the mass deficit or the mass defect,
ΔM.
Mass defect = ΔMD = (mn + mp) - MC
= ( 1.008665 + 1.007825) – 2.014102 = 0.002388 amu
BE = ΔMD c2 = 0.002388 x 931.5
= 2.224 MeV
BE per nucleon = 2. = 1.112 MeV /nucleon
21. Mass defect nucleus = ΔMF-56 = 55.93492 - ( 30 x 1.008665 + 26 x 1.007825)
= 55.934942 amu – 56.46340 = - 0.528458 amu
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