Question 8
In the decay of a
nucleus of 21084 Po, an α-particle is emitted with energy 5.3 MeV.
The emission is
represented by the nuclear equation
21084 Po - -
> ABX
+ α +
energy
(a)
(i) On Fig. 7.1, complete the number and name of the particle, or
particles, represented by A and B in the nuclear equation.
number name
of particle or particles
A
B
Fig. 7.1
[1]
(ii)
State the form of energy given to the α-particle in the decay
of 21084 Po. [1]
(b)
A sample of polonium 21084 Po
emits 7.1 × 1018 α-particles
in one day.
Calculate the mean
power output from the energy of the α-particles. [2]
Reference: Past Exam Paper – November 2014 Paper 22 Q7
Solution:
(a)
(i)
A: 206,
nucleon(s) or neutron(s) and proton(s)
B: 82,
proton(s)
{Alpha
particle is: 42α
So, 210 = A + 4 giving A = 206
84 = B + 2 gibing B = 82}
(ii)
kinetic / EK / KE
(b)
{Power =
Energy / Time
To find
the power, we need to first find the total energy emitted.
When 1
nucleus decays, it emits 5.3 MeV.
Each alpha
particle is emitted from one nucleus.
1 MeV = 1.6×10-13 J}
energy =
5.3 × 1.6×10-13 (J) [=
8.48×10-3 (J)]
{So, 1
nucleus emits 5.3 × 1.6×10-13 J.
7.1×1018 nuclei emit (7.1×1018 × 5.3 × 1.6×10-13) J of energy.
Power =
Energy / Time
Time = 24
h = 24 × 3600 s}
power =
(7.1×1018 × 5.3 × 1.6×10-13) / (3600 × 24)
power = 70 (69.7) W
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