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.}
where did the 2 go in A bit
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