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