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PHYSICS, M8 2023 HSC 26

Consider the following nuclear reaction

\({ }_{\ \ 6}^{12} \text{C} +{ }_1^1 \text{H} \rightarrow{ }_5^9 \text{B} +{ }_2^4 \text{He}\)

The masses of the isotopes in this process are shown in the table.

Isotope Mass (\(u\))
\({ }_{\ \ 6}^{12} \text{C}\) 12.064
\({ }_5^9 \text{B}\) 9.013
\({ }_2^4 \text{He}\) 4.003
\({ }_1^1 \text{H}\) 1.008
Calculate the energy released in this reaction.  (3 marks)

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\(52.164\) \(\text{MeV}\)  or  \(8.357 \times 10^{-12}\) \(\text{J}\)

Show Worked Solution

\(\text{Mass defect}\) \(=m_r-m_p\)  
  \(=(12.064+1.008)-(9.013+4.003)\)  
  \(=0.056\) \(\text{u}\)  

 

\(\text{Energy released}\) \(= 0.056 \times 931.5\) \(\text{MeV}\)  
  \(=52.164\) \(\text{MeV}\)  
  \(=8.357 \times 10^{-12} \) \(\text{J}\)  

PHYSICS, M8 2023 HSC 13 MC

Nucleus \(X\) has a greater binding energy than nucleus \(Y\).

What can be deduced about \(X\) and \(Y\) ?

  1. \(X\) is more stable than \(Y\).
  2. \(Y\) is more stable than \(X\).
  3. \(X\) has a greater mass defect than \(Y\).
  4. \(Y\) has a greater mass defect than \(X\).
Show Answers Only

\(C\)

Show Worked Solution

→ Binding energy is the energy required to split a nucleus into its nucleons. 

→ The mass of the nucleus is less than the sum of the nucleons. The mass equivalent of the binding energy is equal to the mass defect.

→ As nucleus \(X\) has a greater binding energy than nucleus \(Y\), it can be deduced that it also has a greater mass defect.

\(\Rightarrow C\)

♦ Mean mark 43%.

PHYSICS, M8 EQ-Bank 22

Einstein's equation `E = mc^(2)`  is one of the most important equations in the history of physics.

Justify this statement.   (7 marks)

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`E=mc^(2)` at its most basic level, expresses the concept that mass and energy are interchangeable.

Its importance is due to its critical role in explaining and analysing a broad range of ideas, processes and phenomena. These include:

→  The energy sourced from the processes of nuclear fission and fusion, including the energy associated with radioactive decay.

→ The relationship between the binding energy of atoms and their mass defect.

→ Mass dilation of objects approaching the speed of light.

→ The fundamental principles and technology upon which nuclear bombs and nuclear reactors operate.

→ Processes which source energy in stars through the conversion of mass into energy.

→ Processes which allow for particle accelerators to operate, allowing us to investigate into the fundamental structure and properties of matter.

Other equations, such as Newton’s Universal Law of Gravity, do not have the same myriad of applications in fields that make up our current understanding of physics.

It is therefore justified in being called one of the most important equations in the history of physics

Show Worked Solution

`E=mc^(2)` at its most basic level, expresses the concept that mass and energy are interchangeable.

Its importance is due to its critical role in explaining and analysing a broad range of ideas, processes and phenomena. These include:

→  The energy sourced from the processes of nuclear fission and fusion, including the energy associated with radioactive decay.

→ The relationship between the binding energy of atoms and their mass defect.

→ Mass dilation of objects approaching the speed of light.

→ The fundamental principles and technology upon which nuclear bombs and nuclear reactors operate.

→ Processes which source energy in stars through the conversion of mass into energy.

→ Processes which allow for particle accelerators to operate, allowing us to investigate into the fundamental structure and properties of matter.

Other equations, such as Newton’s Universal Law of Gravity, do not have the same myriad of applications in fields that make up our current understanding of physics.

It is therefore justified in being called one of the most important equations in the history of physics

PHYSICS, M8 EQ-Bank 7 MC

The following equation describes the natural decay process of uranium-238.

`\ _(92)^(238)U \rarr\  \ _(90)^(234)Th+ \ _(2)^(4)He`

Which row of the table describes the changes in total mass and total binding energy in the decay of uranium-238?
 

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`A`

Show Worked Solution

→ Mass is converted into energy in this nuclear reaction, so total mass decreases.

→ When elements heavier than iron undergo nuclear fission, the daughter nuclei produced are more stable with a greater binding energy per nucleon. As there are the same number of nucleons before and after the decay, total binding energy increases.

`=>A`

PHYSICS, M8 2019 HSC 36

A radon-198 atom, initially at rest, undergoes alpha decay. The masses of the atoms involved are shown in atomic mass units `(u)`.  
 

The kinetic energy of the polonium atom produced is  `2.55 × 10^(-14)` J.

By considering mass defect, calculate the kinetic energy of the alpha particle, and explain why it is significantly greater than that of the polonium atom.   (7 marks)

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`Delta m` `=197.999-(193.988+4.000260)`  
  `=0.0084` u  
  `=0.0084 xx1.661 xx10^(-27)` (Converting to kg)
  `=1.395 xx10^(-29)` kg  

 

`E` `=mc^2`  
  `=1.395 xx10^(-29)xx(3xx10^(8))^(2)`  
  `=1.256 xx10^(-12)` J  

 
Applying the law of conservation of energy:

`KE_(text{alpha})` `=E_(text{total})-KE_(polonium)`  
  `=1.256 xx10^(-12)-2.44 xx10^(-14)`  
  `=1.23 xx10^(-12)` J  

 
→ As the radon atom is initially at rest, the initial momentum of this reaction system is zero. So, by applying the principle of conservation of momentum, the decay products must move away from each other with equal and opposite momenta.

→ The alpha particle has a significantly lower mass compared to the polonium atom and therefore has a significantly higher velocity.

→ `KE=(1)/(2)mv^(2)=(1)/(2)mv xx v`  and the momenta, `mv`  are equal for both the alpha particle and polonium atom. The alpha particle therefore has a significantly greater kinetic energy due to its greater velocity.

Show Worked Solution
`Delta m` `=197.999-(193.988+4.000260)`  
  `=0.0084` u   
  `=0.0084 xx1.661 xx10^(-27)` (Converting to kg)
  `=1.395 xx10^(-29)` kg  

 

`E` `=mc^2`  
  `=1.395 xx10^(-29)xx(3xx10^(8))^(2)`  
  `=1.256 xx10^(-12)` J  

 
Applying the law of conservation of energy:

`KE_(text{alpha})` `=E_(text{total})-KE_(polonium)`  
  `=1.256 xx10^(-12)-2.44 xx10^(-14)`  
  `=1.23 xx10^(-12)` J  

 
→ As the radon atom is initially at rest, the initial momentum of this reaction system is zero. So, by applying the principle of conservation of momentum, the decay products must move away from each other with equal and opposite momenta.

→ The alpha particle has a significantly lower mass compared to the polonium atom and therefore has a significantly higher velocity.

→ `KE=(1)/(2)mv^(2)=(1)/(2)mv xx v`  and the momenta, `mv`  are equal for both the alpha particle and polonium atom. The alpha particle therefore has a significantly greater kinetic energy due to its greater velocity.


♦♦♦ Mean mark 30%.

PHYSICS, M8 2019 HSC 34

Use the following information to answer this question.
 

Describe both the production and radiation of energy by the sun. In your answer, include a quantitative analysis of both the power output and the surface temperature of the sun.   (9 marks)

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Production of the sun’s energy is predominantly done through the proton-proton chain. Four protons react in the sun’s core to produce one helium atom. The mass of one helium atom is less than the mass of four protons so mass is converted into energy through Einstein’s mass-energy equivalence, `E=mc^2.`

The sun acts as a black body and so radiates its energy in the form of black body radiation. It radiates energy as light with a variety of wavelengths, as shown in the black body curve above. It peaks at a specific wavelength which can be used to calculate its temperature:

`lambda_(max)=5.0xx10^-7` m  (see graph above)

`lambda_(max)` `=(b)/(T)`  
`T` `=(2.898xx10^-3)/(5.0xx10^-7)`  
  `=5796` K  
  `=5800` K  

 
Energy radiated from the sun spreads out over a sphere as it travels away from the sun. The intensity of radiated energy decreases at distances further from the sun, consistent with the inverse square law.

Using the intensity of the sun’s radiation at earth, its power output can be calculated:

`P` `=IA`  
  `=Ixx4pi r^2`  
  `=1360 xx4xx3.142 xx(1.5 xx10^(11))^(2)`  
  `=3.85 xx10^(26)` W  
Show Worked Solution

Production of the sun’s energy is predominantly done through the proton-proton chain. Four protons react in the sun’s core to produce one helium atom. The mass of one helium atom is less than the mass of four protons, so mass is converted into energy through Einstein’s mass-energy equivalence, `E=mc^2.`

The sun acts as a black body and so radiates its energy in the form of black body radiation. It radiates energy as light with a variety of wavelengths, as shown in the black body curve above. It peaks at a specific wavelength which can be used to calculate its temperature:

`lambda_(max)=5.0xx10^-7` m  (see graph above)

`lambda_(max)` `=(b)/(T)`  
`T` `=(2.898xx10^-3)/(5.0xx10^-7)`  
  `=5796` K  
  `=5800` K  

 
Energy radiated from the sun spreads out over a sphere as it travels away from the sun. The intensity of radiated energy decreases at distances further from the sun, consistent with the inverse square law.

Using the intensity of the sun’s radiation at earth, its power output can be calculated:

`P` `=IA`  
  `=Ixx4pi r^2`  
  `=1360 xx4xx3.142 xx(1.5 xx10^(11))^(2)`  
  `=3.85 xx10^(26)` W    

♦♦♦ Mean mark 38%.

PHYSICS M8 2022 HSC 16 MC

The binding energy of helium-4 (He-4) is 28.3 MeV and the binding energy of beryllium-6 (Be-6) is 26.9 MeV.

Which of the following rows in the table is correct?
 

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`A`

Show Worked Solution

The binding energy of a nucleus is the energy required to separate it into individual particles.

→ He-4 requires more energy to separate into individual protons and neutrons.

→ He-4 has 4 nucleons while Be-6 has 6 nucleons.

→ He-4 is less massive.

`=>A`

PHYSICS, M8 2019 HSC 19 MC

Consider the following nuclear reaction.

W  `+`  X  `→`  Y  `+`  Z

Information about W, X and Y is given in the table.
 

Which of the following is a correct statement about energy in this reaction?

  1. The reaction gives out energy because the mass defect of Y is greater than that of either W or X.
  2. It cannot be deduced whether the reaction releases energy because the properties of Z are not known.
  3. The reaction requires an input of energy because the mass defect of the products is greater than the sum of the mass defects of the reactants.
  4. Energy is released by the reaction because the binding energy of the products is greater than the sum of the binding energies of the reactants.
Show Answers Only

`D`

Show Worked Solution

Energy required to break W and  X  into their constituent nucleons:

 → 2.22 `+` 8.48 `=` 10.70 MeV

Energy released in the formation of the products is given by the sum of binding energies of  Y  and  Z:

 → 28.30 MeV ` +` binding energy of Z
   
As this is greater than the sum of binding energies of the reactants regardless of the binding energy of  Z , there is a net release of energy in the reaction.

`=>D`


♦ Mean mark 31%.

PHYSICS, M8 2021 HSC 35

A spacecraft is powered by a radioisotope generator. Pu-238 in the generator undergoes alpha decay, releasing energy. The decay is shown with the mass of each species in atomic mass units, `u`

\begin{array} {ccccc}
\ce{^{238}Pu} & \rightarrow & \ce{^{234}U} & + & \alpha \\
238.0495\ u &  & 234.0409\ u &  & 4.0026\ u \end{array}

  1. Show that the energy released by one decay is  `9.0 × 10^(-13)` J.   (3 marks)
  1. At launch, the generator contains  `9.0 × 10^24`  atoms of Pu-238. The half-life of Pu-238 is 87.7 years.
  2. Calculate the total energy produced by the generator during the first ten years after launch.  (3 marks)
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  1. `9.0xx10^(-13)` J
  2. `6.2xx10^(11)` J
Show Worked Solution

a.   Energy released

`Delta m` `=(234.0409+4.0026)-238.0495=-0.006 u`  
`Delta m` `=0.006 xx1.661 xx10^(-27)=9.966 xx10^(-30)` kg  

  
Using  `E=mc^2` :

`E_text{released}=9.966 xx10^(-30)xx(3xx10^(8))^(2)`

`:. E_text{released}=9.0 xx10^(-13)` J

   

b.   Total energy

`lambda=(ln 2)/(t_((1)/(2)))=(ln 2)/(87.7)=0.0079\ text{year}^(-1)`

`N=N_(0)e^(-lambda t)=9xx10^(24)xxe^(-0.0079 xx10)=8.316 xx10^(24)`

`Delta N=9xx10^(24)-8.316 xx10^(24)=6.84 xx10^(23)`

`E=6.84 xx10^(23)xx9.0 xx10^(-13)`

`:.E=6.2 xx10^(11)` J


♦ Mean mark part (b) 51%.