(a) Explain the following, illustrating your answer with one example in each case: (i) nuclear fusion: (ii) nuclear fission: (iii) radiation hazards.
(b) State two advantages of fusion over fission and explain briefly why, in spite of these advantages, fusion is not normally used for the generation of power.
(c) The current, I in an a.c. circuit is given by the equation: \(I = 30 sin 100\pi t\), where t is the time in seconds. Deduce the following from this equation: (i) frequency of the current (ii) peak value of the current, (iii) r.m.s value of the current.
Explanation
(a)(i) Fusion: Fusion is the coming together of two light nuclei to form a heavier nucleus with the release of energy.
Example: \(^{2} _{1} H + ^{2} _{1} H \to ^{4} _{2} He + Energy\)
(ii) Fission: Fission is the breaking up of heavy nucleus into two nuclei of comparable masses when bombarded with fast neutrons or energetic particles releasing energy in the process.
Example: \(^{235} _{92} U + ^{1} _{0} N \to ^{90} _{39} Kr + ^{144} _{50} Ba + ^{1} _{0} N + ^{1} _{0} N\)
(iii) Radiation Hazards. Radioactive substances can be dangerous if one is exposed to their radiations for a long time. Some of the radiations are highly penetrating and if the body is over-exposed to them, they can destroy the cells in tissues and upset the natural chemistry of the body. Examples include: – Leukaemia – Skin burns
(b) Advantages of fusion over fission include: (1) Easily achieved with lightest element so that nuclear repulsion is easily overcome as nuclei approach each other (2) Raw materials are cheaply available. (3) Produces less dangerous by products. (4) There is no upper limit to the mass of hydrogen that can be obtained. Very high temperature is required to overcome the coulomb repulsive forces between the two light nuclei. This poses severe technological problem due to the fact that materials to withstand this high temperature are difficult to come by.
(c) \(I = I_{m} \sin 2 \pi ft ... (1)\)
I - Instantaneous value of current
\(I_{m}\) - Peak value of current
F - frequency of current
\(I = 30 \sin 100 \pi t ... (2)\)
Comparing (1) and (2),
(i) \(2 \pi f = 100\pi \therefore f = 50 Hz\)
(ii) \(I_{m} = 30A\)
(iii) \(I_{r.m.s} = \frac{I_{m}}{\sqrt{2}}\)
= \(\frac{30}{\sqrt{2}} = 21.2A\)