*Question*

a. One possible fission reaction of uranium-235 (U-235) is

\[{}_{92}^{235}{\rm{U}} + {}_0^1{\rm{n}} \to {}_{56}^{144}{\rm{Ba}} + {}_{36}^{89}{\rm{Kr}} + 3{}_0^1{\rm{n}}\]

The following data are available.

Mass of one atom of U-235 = 235 u

Binding energy per nucleon for U-235 = 7.59 MeV

Binding energy per nucleon for Xe-140 = 8.29 MeV

Binding energy per nucleon for Sr-94 = 8.59 MeV

i. State what is meant by binding energy of a nucleus.[1]

ii. Outline why quantities such as atomic mass and nuclear binding energy are often expressed in non-SI units. [1]

iii. Show that the energy released in the reaction is about 180 MeV. [1]

b. A nuclear power station uses U-235 as fuel. Assume that every fission reaction of U-235 gives rise to 180 MeV of energy.

- Estimate, in J kg–1, the specific energy of U-235. [2]

ii. The power station has a useful power output of 1.2 GW and an efficiency of 36 %. Determine the mass of U-235 that undergoes fission in one day. [2]

c. A sample of waste produced by the reactor contains 1.0 kg of strontium-94 (Sr-94). Sr-94 is radioactive and undergoes beta-minus (β–) decay into a daughter nuclide X. The reaction for this decay is

\(_{{\mkern 1mu} {\mkern 1mu} 38}^{94}{\text{Sr}} \to _{X} + _{ – 1}^{\,\,\,0}{\text{e}} + {\overline {\text{V}} _{\text{e}}}\)

i. Write down the proton number of nuclide X.[1]

ii. The graph shows the variation with time of the mass of Sr-94 remaining in the sample.

State the half-life of Sr-94.[1]

iii. Calculate the mass of Sr-94 remaining in the sample after 10 minutes.[2]

**Answer/Explanation**

### Ans:

a.i. energy required to «completely» separate the nucleons

OR

energy released when a nucleus is formed from its constituent nucleons

aii. The values «in SI units» would be very small

aiii. 140×8.29 + 94x 8.59 − 235x 7.59OR 184 «MeV

b.i. see − «energy »180 x10^{6}x 1.60x 10^{-19} AND «mass =» 235x 1.66x 10^{-27}

= 7.4x 10^{13} «J kg^{-1}

b.ii. energy produced in one day =1.2×10^{9}x24x3600/0.36 = 2.9×10^{14} j

mass = 2.9×10^{14}/7.4×10^{13}=3.9 kg

c.i. 39

c.ii. 75 s

c.iii.

ALTERNATIVE 1

10 min = 8 t_{1/2}

mass remaining =1.0x(1/2)^{8}=3.9×10^{-3} kg

ALTERNATIVE 2

decay constant =ln2/75 =9.24×10^{-3} s^{-1}

mass remaining =1.0xe^{-9.24×10-3×600}=3.9×10^{-3} kg

*Question*

Rhodium-106 (\(_{\,\,\,45}^{106}{\text{Rh}}\)) decays into palladium-106 (\(_{\,\,\,46}^{106}{\text{Pd}}\)) by beta minus (*β*^{–}) decay.

The binding energy per nucleon of rhodium is 8.521 MeV and that of palladium is 8.550 MeV.

*β*^{–} decay is described by the following incomplete Feynman diagram.

a.

Rutherford constructed a model of the atom based on the results of the alpha particle scattering experiment. Describe this model.[2]

State what is meant by the binding energy of a nucleus.[1]

Show that the energy released in the *β*^{–} decay of rhodium is about 3 MeV.[1]

Draw a labelled arrow to complete the Feynman diagram.[1]

Identify particle V.[1]

**Answer/Explanation**

## Markscheme

a.

**«**most of**» **the mass of the atom is confined within a very small volume/nucleus

**«**all**» **the positive charge is confined within a very small volume/nucleus

electrons orbit the nucleus **«**in circular orbits**»***[2 marks]*

the energy needed to separate the nucleons of a nucleus

*OR*

energy released when a nucleus is formed from its nucleons

*Allow neutrons **AND **protons for nucleons*

*Don’t allow constituent parts**[1 mark]*

*Q* = 106 × 8.550 − 106 × 8.521 = 3.07 **«**MeV**»**

**«***Q *≈ 3 Me V**»***[1 mark]*

line with arrow as shown labelled anti-neutrino/\(\bar v\)

*Correct direction of the “arrow” is essential*

*The line drawn must be “upwards” from the vertex in the time direction i.e. above the horizontal*

*[1 mark]*

V = W^{– }^{[1 mark]}

*Question*

a.

A particular K meson has a quark structure \({\rm{\bar u}}\)s. State the charge on this meson.[1]

The Feynman diagram shows the changes that occur during beta minus (β^{–}) decay.

Label the diagram by inserting the **four** missing particle symbols.[2]

Carbon-14 (C-14) is a radioactive isotope which undergoes beta minus (β^{–}) decay to the stable isotope nitrogen-14 (N-14). Energy is released during this decay. Explain why the mass of a C-14 nucleus and the mass of a N-14 nucleus are slightly different even though they have the same nucleon number.[2]

**Answer/Explanation**

## Markscheme

a.

charge: –1«e» * or* negative

*K*

**or**^{−}

*Negative signs required.*

correct symbols for both missing quarks

exchange particle and electron labelled W * or* W

^{–}and e

*e*

**or**^{–}

*Do not allow W*

^{+}or e^{+}or β^{+}Allow β or β^{–}decay products include an electron that has mass* OR *products have energy that has a mass equivalent

*mass/mass defect/binding energy converted to mass/energy of decay products*

**OR**

** «so»**

mass C-14 > mass N-14* OR*mass of

*n*> mass of

*p*

mass of

**OR**

*d*> mass of

*u*

*Accept reference to “lighter” and “heavier” in mass.Do not accept implied comparison, eg “C-14 has greater mass”. Comparison must be explicit as stated in scheme.*