# IBDP Physics 7.2 – Nuclear reactions : IB Style Question Bank SL Paper 2

### 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.

1. 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 x106x 1.60x 10-19 AND «mass =» 235x 1.66x 10-27
= 7.4x 1013 «J kg-1

b.ii. energy produced in one day =1.2×109x24x3600/0.36 = 2.9×1014 j

mass = 2.9×1014/7.4×1013=3.9 kg

c.i. 39

c.ii. 75 s

c.iii.

ALTERNATIVE 1
10 min =  8 t1/2

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]

b.i.

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

b.ii.

Show that the energy released in the β decay of rhodium is about 3 MeV.[1]

c.i.

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

c.ii.

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]

b.i.

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]

b.ii.

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

«Q ≈ 3 Me V»[1 mark]

c.i.

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]

c.ii.

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]

b.

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

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

c.

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 or K

Negative signs required.

b.

correct symbols for both missing quarks

exchange particle and electron labelled W or W and e or e
Do not allow W+ or e+ or β+ Allow β or β

c.

decay products include an electron that has mass
OR
products have energy that has a mass equivalent
OR
mass/mass defect/binding energy converted to mass/energy of decay products

«so»

mass C-14 > mass N-14
OR
mass of n > mass of p
OR
mass of 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.