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Tuesday, December 25, 2018

In the decay of a nucleus of 21084 Po, an α-particle is emitted with energy 5.3 MeV.


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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