*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*

The radioactive nuclide beryllium-10 (Be-10) undergoes beta minus (*β–*) decay to form a stable boron (B) nuclide.

The initial number of nuclei in a pure sample of beryllium-10 is N_{0}. The graph shows how the number of remaining **beryllium **nuclei in the sample varies with time.

An ice sample is moved to a laboratory for analysis. The temperature of the sample is –20 °C.

a.

Identify the missing information for this decay.

[1]

On the graph, sketch how the number of **boron **nuclei in the sample varies with time.[2]

After 4.3 × 10^{6} years,

\[\frac{{{\text{number of produced boron nuclei}}}}{{{\text{number of remaining beryllium nuclei}}}} = 7.\]

Show that the half-life of beryllium-10 is 1.4 × 10^{6} years.[3]

Beryllium-10 is used to investigate ice samples from Antarctica. A sample of ice initially contains 7.6 × 10^{11} atoms of beryllium-10. State the number of remaining beryllium-10 nuclei in the sample after 2.8 × 10^{6} years.[1]

State what is meant by thermal radiation.[1]

Discuss how the frequency of the radiation emitted by a black body can be used to estimate the temperature of the body.[2]

Calculate the peak wavelength in the intensity of the radiation emitted by the ice sample.[2]

Derive the units of intensity in terms of fundamental SI units.[2]

**▶️Answer/Explanation**

## Markscheme

a.

\(_{{\mkern 1mu} {\mkern 1mu} 4}^{10}{\text{Be}} \to _{{\mkern 1mu} {\mkern 1mu} 5}^{10}{\text{B}} + \beta + {\overline {\text{V}} _{\text{e}}}\)

conservation of mass number ** AND **charge \(_{\,\,5}^{10}{\text{B}}\), \(_{{\mkern 1mu} {\mkern 1mu} 4}^{10}{\text{Be}}\)

*Correct identification of both missing values required for **[1]**.***[1 mark]**

correct shape *ie *increasing from 0 to about 0.80 N_{0}

crosses given line at 0.50 N_{0}

**[2 marks]**

*ALTERNATIVE 1*

fraction of Be = \(\frac{1}{8}\), 12.5%, or 0.125

therefore 3 half lives have elapsed

\({t_{\frac{1}{2}}} = \frac{{4.3 \times {{10}^6}}}{3} = 1.43 \times {10^6}\) **«**≈ 1.4 × 10^{6}**»** **«**y**»**

*ALTERNATIVE 2*

fraction of Be = \(\frac{1}{8}\), 12.5%, or 0.125

\(\frac{1}{8} = {{\text{e}}^{ – \lambda }}(4.3 \times {10^6})\) leading to *λ* = 4.836 × 10^{–7} **«**y**»**^{–1}

\(\frac{{\ln 2}}{\lambda }\) = 1.43 × 10^{6} **«**y**»**

*Must see at least one extra sig fig in final answer.**[3 marks]*

1.9 × 10^{11}**[1 mark]**

emission of (infrared) electromagnetic/infrared energy/waves/radiation.**[1 mark]**

the (peak) wavelength of emitted em waves depends on temperature of emitter/reference to Wein’s Law

so frequency/color depends on temperature**[2 marks]**

\(\lambda = \frac{{2.90 \times {{10}^{ – 3}}}}{{253}}\)

= 1.1 × 10^{–5} **«**m**»**

*Allow ECF from MP1 (incorrect temperature).***[2 marks]**

correct units for Intensity (allow *W, Nms ^{–}*

^{1 }

*OR Js*

^{–}^{1 }

*in numerator)*

rearrangement into proper SI units = kgs^{–3}

*Allow ECF for MP2 if final answer is in fundamental units.**[2 marks]*

*Question*

The first scientists to identify alpha particles by a direct method were Rutherford and Royds. They knew that radium-226 (\({}_{86}^{226}{\text{Ra}}\)) decays by alpha emission to form a nuclide known as radon (Rn).

a.

Write down the missing values in the nuclear equation for this decay.

[1]

Rutherford and Royds put some pure radium-226 in a small closed cylinder A. Cylinder A is fixed in the centre of a larger closed cylinder B.

At the start of the experiment all the air was removed from cylinder B. The alpha particles combined with electrons as they moved through the wall of cylinder A to form helium gas in cylinder B.

The wall of cylinder A is made from glass. Outline why this glass wall had to be very thin.[1]

Rutherford and Royds expected 2.7 x 10^{15} alpha particles to be emitted during the experiment. The experiment was carried out at a temperature of 18 °C. The volume of cylinder B was 1.3 x 10^{–5} m^{3} and the volume of cylinder A was negligible. Calculate the pressure of the helium gas that was collected in cylinder B.[3]

Rutherford and Royds identified the helium gas in cylinder B by observing its emission spectrum. Outline, with reference to atomic energy levels, how an emission spectrum is formed.[3]

The work was first reported in a peer-reviewed scientific journal. Outline why Rutherford and Royds chose to publish their work in this way.[1]

**▶️Answer/Explanation**

### Markscheme

a.

222 * AND *4

*Both needed.*

alpha particles highly ionizing**OR**

alpha particles have a low penetration power**OR**

thin glass increases probability of alpha crossing glass**OR**

decreases probability of alpha striking atom/nucleus/molecule

conversion of temperature to 291 K

*p* = 4.5 x 10^{–9} x 8.31 x «\(\frac{{2.91}}{{1.3 \times {{10}^{ – 5}}}}\)»

**OR**

*p* = 2.7 x 10^{15} x 1.38 x 10^{–23} x «\(\frac{{2.91}}{{1.3 \times {{10}^{ – 5}}}}\)»

0.83 or 0.84 «Pa»

electron/atom drops from high energy state/level to low state

energy levels are discrete

wavelength/frequency of photon is related to energy change * or *quotes

*E*=

*hf*= \(\frac{{hc}}{\lambda }\)

**or**Eand is therefore also discrete

peer review guarantees the validity of the work**OR**

means that readers have confidence in the validity of work

*OWTTE*