One nuclear reaction that can take place in a nuclear reactor may be represented, in part, by the equation

Question 10

One nuclear reaction that can take place in a nuclear reactor may be represented, in part, by the equation

Data for a nucleus and some particles are given in Fig. 12.1.

Fig. 12.1

(a) Complete the nuclear reaction shown above. [1]

(b) (i) Show that the energy equivalent to 1.00 u is 934 MeV. [3]

(ii) Calculate the binding energy per nucleon, in MeV, of lanthanum-139 (¹³⁹₅₇ La). [3]

(c) State and explain whether the binding energy per nucleon of uranium-235 (²³⁵₉₂ U) will be greater, equal to or less than your answer in (b)(ii). [3]

[Total: 10]

Reference: Past Exam Paper – June 2017 Paper 42 Q12

Solution:

(a)

7 ⁰_-1e  

{The overall nucleon number and proton number on both sides of the equation should be equal.

Let the particle emitted be: ^A_ZX

Nucleon number: 235 + 1 = 95 + 139 + 2(1) + A

A + 236 = 236

A = 0

Proton number:

92 + 0 = 42 + 57 + 2(0) + Z

Z + 99 = 92

Z = – 7

As, the particle has a nucleon number of zero and a negative proton number. This indicates that it is negatively charged.

So, it can be concluded that it is an electron.

However, since the charge of an electron is – 1, we need to have 7 electrons.}

(b)

(i)

{The equivalent energy associated with some mass can be obtained by Einstein’s equation:}

E = mc²

{From the list of date, 1 u = 1.66 × 10^-27 kg}

E = 1.66 × 10^-27 × (3.00 × 10⁸)²

E = 1.494 × 10^-10 J

{This is the energy in joules. To convert from joule to eV, we divide by the charge of an electron (1.6 × 10^-19). This gives the energy in eV.

To obtain the energy in MeV, we multiply by 10^-6.

So, on the overall, we divide by 1.6 × 10^-13 to convert from joules to MeV.}

division by 1.60 × 10^-13 clear to give 934 MeV

(ii)

{The energy of a lanthanum nucleus (with the protons and neutrons inside) is less than the sum of energies of the individual particles. This is because some of the mass has been converted into (binding) energy to keep the particles together inside the nucleus.

So, there is a loss in mass (mass defect).

Mass defect Δm = mass of individual particles – mass of lanthanum nucleus

As given in the table,

Mass of lanthanum nucleus = 138.955 u

Lanthanum nucleus consists of 57 protons (mass of each = 1.00728 u) and (139 – 57 =) 82 neutrons (mass of each = 1.00863 u).}

Δm = (82 × 1.00863u) + (57 × 1.00728u) – 138.955u

Δm = (–) 1.16762 (u) 

{Equivalent energy of 1 u = 934 MeV. So, energy due to the above mass defect is:}

energy = 1.16762 × 934

{Lanthanum contains 139 nucleons.}

energy per nucleon = (1.16762 × 934) / 139

energy = 7.85 MeV

(c)

{Considering the stability of nuclei,}

Above (proton number) A = 56, the binding energy per nucleon decreases as A increases.

Uranium-235 has a larger nucleon number (than lanthanum-139), so the binding energy per nucleon is less.