(a) (i) What is nuclear fission?
(ii) State the function of each of the following ma-terials in a nuclear fision reactor:
(\(\alpha\)) graphite, (\(\beta\)) boron rods; (\(\gamma\)) liquid sodium
(b) The table below gives some of the energy levels of a hydrogen atom.
| n | 1 | 2 | 3 | 4 | 5 | \(\infty\) |
| E\(_n\)/\_eV\) | -13.60 | -3.39 | -1.51 | -0.85 | -0.54 | 0.00 |
(i) Draw the energy level diagram for the atom.
(ii) Determine the wavelength of the photon emitted when the atom goes from the energy state n = 3 to the ground state. [h = 6.6 x 10\(^{-34}\) Js, c = 3 x 10\(^{8}\)ms\(^{-1}\), e = 1.6 x 10\(^{-19}\)]
(c) A piece of ancient bone from an excavation site showed \(^{14}_{6}\)C activity of 9.5 disintegrations per minute per 1.0x10\(^{-3}\)kg. If a bone specimen from a living creature shows \(^{14}_{6}\)C activity of 12.0 disintegrations per minute per 1.0 x 1 0\(^{-3}\), determine the age of the ancient bone. [Half - life of \(^{14}_{6}\)C = 5572 years].
(a)(i) Nuclear Fission is the splitting up of a heavy nucleus into lighter nuclei with release of a huge amount of energy and neutrons.
(ii) (\(\alpha\)) graphite; (\(\beta\)) Boron rods; (\(\gamma\) )Liquid sodium
|
Material |
Function |
|
Graphite |
Slows down high energetic neutrons in the reaction chamber. |
|
Boron rods |
To absorb excess neutrons so as to control fission rate |
|
Liquid sodium |
Extraction / conduction of excess heat from the reaction chamber. |
(b)(i) E\(_n\), n = \(\alpha\) ---------------------------------- 0.eV
E\(_4\), n = 5 ---------------------------------------------------- - 0.54eV
E\(_3\), n = 4 ---------------------------------------------------- - 0.85eV
E\(_2\), n = 3 ---------------------------------------------------- 1.51eV
E\(_1\), n = 2 ------------------------------------------------------ 3.39eV
(ii) Using \(\lambda\) = \(\frac{hc}{\Delta E}\)
= \(\frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{(-1.51 - ( - 13.60)) \times 1.6 \times 10^{-19}}\)
= 1.02 x 10\(^{-7}\)m
(c) \(\lambda\) = \(\frac{0.693}{ T^{\frac{1}{2}}}\) = \(\frac{0.693}{5572}\)
= 0.0001244 years\(^{-1}\)
N = Noe\(^{- \lambda t}\)
e\(^{0.0001244t}\) = 1.2632
0.0001244t = 0.234
t = 1878 years
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