The isotope iodine-131 (13153I) is radioactive with a decay constant of 8.6 × 10-2 day-1.

Question 13

The isotope iodine-131 (¹³¹₅₃I) is radioactive with a decay constant of 8.6 × 10^-2 day^-1.

β^- particles are emitted with a maximum energy of 0.61 MeV.

(a) State what is meant by

(i) radioactive, [2]

(ii) decay constant. [2]

(b) Explain why the emitted β^- particles have a range of energies. [2]

(c) A sample of blood contains 1.2 × 10^-9 g of iodine-131.

Determine, for this sample of blood,

(i) the activity of the iodine-131, [3]

(ii) the time for the activity of the iodine-131 to be reduced to 1/50 of the activity calculated in (i). [2]

[Total: 11]

Reference: Past Exam Paper – November 2017 Paper 42 Q12

Solution:

(a)

(i) A radioactive nucleus is an unstable nucleus that emits particles / EM radiation / ionizing radiation randomly and spontaneously.

(ii) The decay constant is the probability of decay of a nucleus per unit time.

(b) The emitted beta particle has to share the energy of the disintegration with the emitted antineutrino.

(c)

(i)

{A = λN

We need to find the value of N.

The relative atomic mass of iodine is 131 g.

This mass contains 6.02 × 10²³ nuclei.

131 g - - - > 6.02 × 10²³ nuclei

1.2×10^-9 g - - - > [(1.2 × 10^-9) / 131] × 6.02 × 10²³ nuclei}

number = [(1.2 × 10^-9) / 131] × 6.02 × 10²³

number ( = 5.51 × 10¹²)

A = λN

{Decay constant λ = 8.6×10^-2 day^-1

We need to convert the decay constant from day^-1 to s^-1.

In a day, there are (24 × 3600) seconds.

Decay constant λ = 0.086 / (24 × 3600) s^-1 }

A = [0.086 / (24 × 3600)] × 5.51 × 1012

A = 5.5 × 10⁶ Bq

(ii)

{A = A₀ exp(-λt)

A / A₀ = exp(-λt)                      and A / A₀ = 1 / 50}

1 / 50 = exp(–0.086t)

{1 / 50 = exp(–0.086t)

ln (1/50) = –0.086t }

t = 45 days

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.}