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Q13.5(ii):
The Q value of a nuclear reaction \(A + b \rightarrow C + d\) is defined by \(Q = [ m_{A} + m_{b} – m_{C} – m_{d} ]c^{2}\) where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.
(ii) \(^{12}_{6}C+ ^{12}_{6}C \rightarrow ^{20}_{10}Ne+ ^{4}_{2}He\)
Atomic masses are given to be
\(m (\(^{2}_{1}H\)) = 2.014102 u\)
\(m (\(^{3}_{1}H\)) = 3.016049 u\)
\(m (\(^{12}_{6}C\)) = 12.000000 u\)
\(m (\(^{20}_{10}Ne\)) = 19.992439 u\)
Solution :
Step 1: Identifying Reactant and Product Masses
The nuclear reaction is given as: $^{12}_{6}C + ^{12}_{6}C \rightarrow ^{20}_{10}Ne + ^{4}_{2}He$.
From the problem statement and standard physical constants, we identify the following atomic masses:
- Mass of Reactant 1 ($m_A$): $m(^{12}_{6}C) = 12.000000 \, u$
- Mass of Reactant 2 ($m_b$): $m(^{12}_{6}C) = 12.000000 \, u$
- Mass of Product 1 ($m_C$): $m(^{20}_{10}Ne) = 19.992439 \, u$
- Mass of Product 2 ($m_d$): $m(^{4}_{2}He) = 4.002603 \, u$ (standard value for Alpha particle)
Step 2: Calculating Mass Defect ($\Delta m$)
The mass defect is calculated by finding the difference between the total mass of the reactants and the total mass of the products [Per the definition of binding energy and mass-energy equivalence]:
$\Delta m = [ (m_A + m_b) - (m_C + m_d) ]$
$\Delta m = [ (12.000000 + 12.000000) - (19.992439 + 4.002603) ] \, u$
$\Delta m = [ 24.000000 - 23.995042 ] \, u$
$\Delta m = 0.004958 \, u$
Step 3: Calculating Q-value via Energy Equivalence
The Q-value is given by $Q = \Delta m \times c^{2}$. Given that $1 \, u \approx 931.5 \, MeV/c^{2}$ [By Einstein's Mass-Energy Equivalence Principle]:
$Q = 0.004958 \, u \times 931.5 \, MeV/u$
$Q \approx 4.6183 \, MeV$
Step 4: Classification of Reaction
The reaction type is determined by the sign of the Q-value:
| Condition | Classification |
|---|---|
| $Q > 0$ | Exothermic (Energy released) |
| $Q < 0$ | Endothermic (Energy absorbed) |
Since the calculated $Q = +4.6183 \, MeV$, which is greater than zero, the reaction releases energy to the surroundings.
Final Solution: The Q-value is $4.6183 \, MeV$. Since $Q > 0$, the reaction is exothermic.
More Questions from Class 12 Physics Nuclei EXERCISES
- Q13.1: Obtain the binding energy (in MeV) of a nitrogen nucleus $(_7^{14}N)$, given $m(_7^{14}N) = 14.00307$ u
- Q13.1: Obtain the binding energy (in MeV) of a nitrogen nucleus (\(^{14}_{7}N\)), given \(m (\(^{14}_{7}N\)) =14.00307 u\)
- Q13.10: From the relation $R = R_0 A^{1/3}$, where $R_0$ is a constant and $A$ is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of $A$).
- Q13.10: From the relation \(R = R_{0}A^{\frac{1}{3}}\), where \(R_{0}\) is a constant and \(A\) is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).
- Q13.2: Obtain the binding energy of the nuclei $_26^{56}Fe$ and $_83^{209}Bi$ in units of MeV from the following data: $m(_26^{56}Fe) = 55.934939$ u $m(_83^{209}Bi) = 208.980388$ u
- Q13.2: Obtain the binding energy of the nuclei \(^{56}_{26}Fe\) and \(^{209}_{83}Bi\) in units of MeV from the following data: \(m (\(^{56}_{26}Fe\)) = 55.934939 u\) \(m (\(^{209}_{83}Bi\)) = 208.980388 u\)
- Q13.3: A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of $_29^{63}Cu$ atoms (of mass 62.92960 u).
- Q13.3: A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of \(^{63}_{29}Cu\) atoms (of mass 62.92960 u).
- Q13.4: Obtain approximately the ratio of the nuclear radii of the gold isotope $_79^{197}Au$ and the silver isotope $_47^{107}Ag$.
- Q13.4: Obtain approximately the ratio of the nuclear radii of the gold isotope \(^{197}_{79}Au\) and the silver isotope \(^{107}_{47}Ag\).
- Q13.5(i): The $Q$ value of a nuclear reaction $A + b \rightarrow C + d$ is defined by $Q = [m_A + m_b – m_C – m_d]c^2$ where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic. (i) $_1^1H + _1^3H \rightarrow _1^2H + _1^2H$ Atomic masses are given to be $m(_1^2H) = 2.014102$ u $m(_1^3H) = 3.016049$ u $m(_6^{12}C) = 12.000000$ u $m(_{10}^{20}Ne) = 19.992439$ u
- Q13.5(i): The Q value of a nuclear reaction \(A + b \rightarrow C + d\) is defined by \(Q = [ m_{A} + m_{b} – m_{C} – m_{d} ]c^{2}\) where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic. (i) \(^{1}_{1}H+ ^{3}_{1}H \rightarrow ^{2}_{1}H+ ^{2}_{1}H\) Atomic masses are given to be \(m (\(^{2}_{1}H\)) = 2.014102 u\) \(m (\(^{3}_{1}H\)) = 3.016049 u\) \(m (\(^{12}_{6}C\)) = 12.000000 u\) \(m (\(^{20}_{10}Ne\)) = 19.992439 u\)
- Q13.5(ii): The $Q$ value of a nuclear reaction $A + b \rightarrow C + d$ is defined by $Q = [m_A + m_b – m_C – m_d]c^2$ where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic. (ii) $_6^{12}C + _6^{12}C \rightarrow _{10}^{20}Ne + _2^4He$ Atomic masses are given to be $m(_1^2H) = 2.014102$ u $m(_1^3H) = 3.016049$ u $m(_6^{12}C) = 12.000000$ u $m(_{10}^{20}Ne) = 19.992439$ u
- Q13.6: Suppose, we think of fission of a $_26^{56}Fe$ nucleus into two equal fragments, $_13^{28}Al$. Is the fission energetically possible? Argue by working out $Q$ of the process. Given $m(_26^{56}Fe) = 55.93494$ u and $m(_13^{28}Al) = 27.98191$ u.
- Q13.6: Suppose, we think of fission of a \(^{56}_{26}Fe\) nucleus into two equal fragments, \(^{28}_{13}Al\). Is the fission energetically possible? Argue by working out Q of the process. Given \(m (\(^{56}_{26}Fe\)) = 55.93494 u\) and \(m (\(^{28}_{13}Al\)) = 27.98191 u\).
- Q13.7: The fission properties of $_94^{239}Pu$ are very similar to those of $_92^{235}U$. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure $_94^{239}Pu$ undergo fission?
- Q13.7: The fission properties of \(^{239}_{94}Pu\) are very similar to those of \(^{235}_{92}U\). The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure \(^{239}_{94}Pu\) undergo fission?
- Q13.8: How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as $_1^2H + _1^2H \rightarrow _2^3He + n + 3.27$ MeV
- Q13.8: How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as \(^{2}_{1}H+ ^{2}_{1}H \rightarrow ^{3}_{2}He+n+3.27 MeV\)
- Q13.9: Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)
- Q13.9: Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)
CBSE Solutions for Class 12 Physics Nuclei
Chapters in CBSE - Class 12 Physics
- Electric Charges and Fields
- Electrostatic Potential and Capacitance
- Current Electricity
- Moving Charges and Magnetism
- Magnetism and Matter
- Electromagnetic Induction
- Alternating Current
- Electromagnetic Waves
- Ray Optics and Optical Instruments
- Wave Optics
- Dual Nature of Radiation and Matter
- Atoms
- Semiconductor Electronics
- Communication Systems
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