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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.)
Solution :
Step 1: Identifying Given Variables
To calculate the electrostatic potential energy (potential barrier) between two deuterons at the moment of contact, we define the following parameters:
- Charge of a deuteron ($q_1 = q_2$): $1e = 1.6 \times 10^{-19} \text{ C}$
- Radius of a deuteron ($R$): $2.0 \text{ fm} = 2.0 \times 10^{-15} \text{ m}$
- Distance of closest approach ($r$): When two spheres of radius $R$ touch, the distance between their centers is $r = R + R = 2R = 4.0 \times 10^{-15} \text{ m}$
- Coulomb's constant ($k$): $8.987 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2$ (approximated as $9 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2$)
Step 2: Applying the Coulomb Potential Formula
The potential barrier height ($V$) is equal to the electrostatic potential energy of the system at the distance $r$, according to the principle of conservation of energy [Per Coulomb's Law for point charges]:
$V = \frac{1}{4\pi\epsilon_0} \cdot \frac{q_1 q_2}{r}$
Step 3: Algebraic Substitution & Calculation
Substituting the known values into the equation:
$V = \frac{(9 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2) \times (1.6 \times 10^{-19} \text{ C})^2}{4.0 \times 10^{-15} \text{ m}}$
Calculating the numerator:
$(1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \text{ C}^2$
$9 \times 10^9 \times 2.56 \times 10^{-38} = 23.04 \times 10^{-29} \text{ J}\cdot\text{m}$
Dividing by the separation distance:
$V = \frac{23.04 \times 10^{-29}}{4.0 \times 10^{-15}} = 5.76 \times 10^{-14} \text{ J}$
Step 4: Unit Conversion to Electron-Volts (eV)
To express the energy in standard nuclear units, we divide by $1.6 \times 10^{-19} \text{ J/eV}$:
$V(\text{eV}) = \frac{5.76 \times 10^{-14} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} = 3.6 \times 10^5 \text{ eV} = 0.36 \text{ MeV}$
Visual Representation of the Collision
Final Solution: The height of the potential barrier for a head-on collision of two deuterons is 0.36 MeV.
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
- Q13.5(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\)
- 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.)
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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