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YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

Tuesday, February 26, 2019

One possible nuclear reaction that takes place in a nuclear reactor is given by the equation


Question 12
One possible nuclear reaction that takes place in a nuclear reactor is given by the equation



Data for the nuclei and particles are given in Fig. 12.1.


Fig. 12.1

(a) Determine, for this nuclear reaction, the value of x. [1]

(b) (i) Show that the energy equivalent to 1.00 u is 934 MeV. [3]
(ii) Calculate the energy, in MeV, released in this reaction. Give your answer to three
significant figures. [3]

(c) Suggest the forms of energy into which the energy calculated in (b)(ii) is transformed. [2]
[Total: 9]





Reference: Past Exam Paper – June 2017 Paper 41 & 43 Q12





Solution:
(a)
{Atomic number and nucleon number should be conserved.
Consider atomic number:
92 + 0 = 42 + 57 + 2(0) + x(-1)
92 = 99 – x
x = 99 – 92 = 7}

x = 7


(b) (i)
E = mc2             
{1 u = 1.66 × 10-27 kg}
E = 1.66 × 10-27 × (3.0 × 108)2
E = 1.494 × 10-10 J

{To obtain energy in eV, we divide by 1.6×10-19.
To obtain energy in MeV, we further divide by 106.
So, to convert energy from J to MeV, we divide by 1.6×10-13.}

division by 1.6 × 10-13 clear to give 934 MeV

(ii)
{Energy of reactants = Energy of products + Energy released
Energy released = Energy of reactants – Energy of products

First we find the change in mass.}
Δm = (235.123 + 1.00863) – (94.945 + 138.955 + 2×1.00863 + 7×(5.49 × 10-4))
or
Δm = 235.123 – (94.945 + 138.955 + 1×1.00863 + 7×(5.49 × 10-4))

Δm = 0.21053 u

{As shown above, the energy equivalent to 1.00 u is 934 MeV.
So, we find the energy equivalent of the change in mass found.}

energy = 0.21053 × 934
energy = 197 MeV


(c)
kinetic energy of nuclei/particles/products/fragments
γ–ray photon energy

{Nuclear reactions transform energy to the kinetic energy of fission fragments and also gamma
radiation.}

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