Students can find the solutions of the Problems of chapter 6.6 - Nuclear Reaction on this page. Basically, nuclear reactions are processes in which one or more nuclides are produced from the collisions between two atomic nuclei or one atomic nucleus and a subatomic particle. Nuclear fission, nuclear fusion, alpha decay, gamma emission etc., are the main concepts in this chapter. Students are recommended to go through these solutions so that they can be familiar with the type of questions from Nuclear Reaction.
1: An alpha-particle with kinetic energy T_{α} = 7 .0 MeV is scattered elastically by an initially stationary Li^{6} nucleus. Find the kinetic energy of the recoil nucleus if the angle of divergence of the two particles is ϕ = 60°.
Solution:
Initial momentum of a particle is √(2mT_{α}) i
Where i is a unit vector in the incident direction
Final momenta are respectively
Squaring P^{2}_{α} + P^{2}_{Li} + 2P_{α} P_{Li }cos ϕ = 2m T_{α} ……(1)
Also, by energy conservation, P^{2}_{α} /2m + P^{2}_{Li}/2M = T_{α}
P^{2}_{α} + m/M P^{2}_{Li} = 2mT_{α} …..(2)
Subtracting (2) from (1)
P_{Li}[(1-m/M)P_{Li} + 2P_{ α} cos ϕ] = 0
If P_{Li } = 0, then P_{α} = (-1/2)(1 – m/M)P_{Li} sec ϕ
Since P_{α} and P_{Li } are both positive number (being magnitudes of vectors)
we must have, -1 ≤ cos ϕ < 0 if m < M.
This being understood, we write
(P^{2}_{Li}/2M) [1 + M/4m(1-m/M)^{2} sec^{2} ϕ] = T_{α}
(P^{2}_{Li}/2M) = T_{α}/[1 + M/4m(1-m/M)^{2} sec^{2} ϕ]
As ϕ is given. If we take ϕ = 120^{0}, we have recoil energy of Li = 6 MeV.
2: A neutron collides elastically with an initially stationary deuteron. Find the fraction of the kinetic energy lost by the neutron
(a) in a heads on collision;
(b) in scattering at right angles.
Solution:
(a) In a head on collision
√(2mT) = p_{d} + p_{n}
T = p^{2}_{d}/2M + p^{2}_{n}/2m
Where p_{d} and p_{n} are the momenta of deuteron and neutron after the collision. Squaring ,2mT = p^{2}_{d} + p^{2}_{n} + 2p_{d}p_{n}
p^{2}_{n} + (m/M)p^{2}_{d} = 2mT
or since p_{d} = 0 in a head on collisions
p_{n}=(-1/2)(1-m/M)p_{d}
Going back to energy conservation
(p^{2}_{d}/2M) (1+ M/4m (1 – m/M)^{2}) = T
So, p_{d}^{2}/2M = [4mM/(m+M)^{2}] T
This is the energy lost by neutron. So, the fraction of energy lost is
Η = 4mM/(m+M)^{2} = 8/9
(b) In this case neutron is scattered by 90^{o}. Then we have
Then by energy conservation:
Or (p_{n}^{2}/2m)(1 + m/M) = T(1 - m/M)
Or T - p_{n}^{2}/2m) = 2m/(m+M)T
Or fraction of energy lost is η = 2m/(m+M)T = 2/3
3: Find the greatest possible angle through which a deuteron is scattered as a result of elastic collision with an initially stationary proton.
Solution:
From conservation of momentum:
or p_{p}^{2} = 2MT + p_{d}^{2} - 2√(2MT) p_{d} cosθ
From energy conservation
T = p_{p}^{2} /2m + p_{d}^{2}/2M
(M = mass of denteron and m = mass of proton)
so, p_{p}^{2} = 2mT – (m/M) p_{d}^{2}
Therefore, p_{d}^{2}(1 + m/M) - 2 √(2MT) p_{d} cos θ + 2(M-m)T = 0
For real roots, 4(2MT)cos^{2} θ - 4x2(M-m)T(1 + m/M) ≥0
Cos^{2} θ ≥ (1 -m^{2}/M^{2})
Hence, sin^{2} θ ≤ m^{2}/M^{2}
i.e. θ ≤ sin^{-1}(m/M)
For deuteron-proton scaltering, θ_{max} = 30^{o}.
4: Assuming the radius of a nucleus to be equal to 0.13 A^{1/3} pm, where A is its mass number, evaluate the density of nuclei and the number of nucleons per unit volume of the nucleus.
Solution:
This problem has a misprint, actually the radius R of a nucleus is given as 1.3 A^{1/3} pm where f_{m} = 10^{-15} m
Then the number of nucleous per unit volume is
A/[(4π/3) R^{3}] = 1.09 x 10^{38} per cc
The corresponding mass density is (1.09 x 10^{-38} x mass of a nucleon) per cc = 1.82 x 10^{11 }kg/cc.
5: Write missing symbols, denoted by x, in the following nuclear reactions:
(a) B^{10} (x, α) Be^{8} (b) O^{17} (d, n) x; (c) Na^{23} (p, x) Ne^{20}; (d) x (p, n) Ar^{37}.
Solution:
(a) The particle x must carry two nucleons and a unit of positive charge.
The reaction is B^{10}(d, α)Be^{β}
(b) The particle x must contain a proton in addition to the constituents of O^{17}. Thus the reaction is O^{17}(d, n)F^{18}
(c) The particle x must carry nucleon number 4 and two units of +ve charge. Thus the particle must be x = α and the reaction is Na^{23}(p, α)Ne^{20}.
(d) The particle x must carry mass number 37 and have one unit less of positive charge. Thus x = Cl^{37} and the reaction is Cl^{37}(p,n) Ar^{37}.
6: Demonstrate that the binding energy of a nucleus with mass number A and charge Z can be found from E_{b} = Z ∆_{H} + (A - Z) ∆_{n} - ∆.
Solution:
From the basic formula, E_{b} = Z ∆_{H} + (A - Z) ∆_{n} - ∆
We define AH = mH - 1 amu
An = - 1 amu
A = M - A amu
Then clearly E_{b} - Z A# + (A - Z ) A_{n} – A
7: Find the binding energy of a nucleus consisting of equal numbers of protons and neutrons and having the radius one and a half times smaller than that of Al^{27} nucleus.
Solution:
The mass number of the given nucleus must be 27/(3/2)^{3} = 8
Thus the nucleus is Be^{8}. Then The binding energy is E_{b} - 4 x 0-00867 + 4 + x 0-00783 – 0-0-00531 amu
= 0-06069 amu = 56-5 MeV
(On using 1 amu = 931 MeV.)
8: Making use of the tables of atomic masses, find: (a) the mean binding energy per one nucleon in O^{16} nucleus; (b) the binding energy of a neutron and an alpha-particle in a B^{11} nucleus; (c) the energy required for separation of an O^{16} nucleus into four identical particles.
Solution:
(a) Total binding energy of the O^{16} nucleus is
E_{b} = 8 x .00867 + 8 x .00783 + 0.00509 amu
= 0.13709 amu = 127.6 MeV
So B.E. per nucleon is 7.98 Mev/nucleon
(b) B.E. of neutron in B^{11} nucleus = B.E. of B^{11} – B.E. of B^{10}
(since on removing a neutron from B^{11} we get B^{10})
=∆_{n} - ∆_{B11} + ∆_{B10} = 0.01231 amu = 11.46 MeV
B.E. of ( an α-particle in B^{11}) = B.E. of B^{1} - B.E. of Li^{7} - B.E. of α
(since on removing an a from B^{11} we get Li^{7} )
= -∆_{B11 }+ ∆_{Li} + ∆_{α}
= - 0.00930 + 0.01601 + 0.00260 = 0.00931 amu = 8.67 MeV
(c) This energy is
[B.E. of O^{16} + 4 (B.E. of a particles)]
= -∆_{0}^{16 }+ 4∆_{α}
= 4 x 0-00260 + 0.00509 = 0.01549 amu = 14.42 MeV
9: Find the difference in binding energies of a neutron and a proton in a B^{11} nucleus. Explain why there is the difference.
Solution:
B.E. o f a neutron in B^{11} - B.E. of a proton in B^{11}
=(∆_{n} - ∆_{B}^{11} + ∆_{B}^{10}) – (∆_{p} - ∆_{B}^{11} + ∆_{Be}^{10})
= 0.00024 amu = 0.223 MeV
The difference in binding energy is essentially due to the coulomb repulsion between the proton and the residual nucleus Be^{10} which together constitute B^{11}.
10: Find the energy required for separation of a Ne^{20} nucleus into two alpha-particles and a C^{12} nucleus if it is known that the binding energies per one nucleon in Ne^{20}, He^{4}, and C^{12} nuclei are equal to 8.03, 7.07, and 7.68 MeV respectively.
Solution:
Required energy is simply the difference in total binding energies
= B.E. of Ne^{20} - 2 (BE. of He^{4}) - B.E. of C^{12}
= 20 ε_{Ne} - 8 ε_{α} – 12 ε_{C}
(ε is binding energy per unit nucleon.)
Substitution gives 11.88MeV.
11: Calculate in atomic mass units the mass of
(a) a Li^{8} atom whose nucleus has the binding energy 41.3 MeV;
(b) a C^{10} nucleus whose binding energy per nucleon is equal to 6.04 MeV.
Solution:
We have for Li^{8}
41.3 MeV = 0.044361 amu = 3Δ_{H} + 5Δ_{n} – Δ
Hence Δ = 3 x 0.00783 + 5 x 0.00867 - 0.09436 - 0.02248 amu
(b) For C^{10}
10 x 6.04 = 60.4 MeV - 0-06488 amu
Hence Δ = 6 x 0.00783 + 4 x 0-00867 - 0.06488 = 0.01678 amu
Hence the mass of C^{10} is 10.01678 amu.
12: The nuclei involved in the nuclear reaction A_{1} + A_{2} → A_{3} + A_{4} have the binding energies E_{1}, E_{2}, E_{3}, and E_{4}. Find the energy of this reaction.
Solution:
Suppose M_{1}, M_{2}, M_{3}, M_{4} are the rest masses of the nuclei A_{1}. lf A_{2} , A_{3} and A_{4} participating in the reaction
A_{1} + A_{2} → A_{3} + A_{4} + Q
Here Q is the energy released. Then by conservation of energy.
Q = c^{2}(M_{1} + M_{2} - M_{3} - M_{4})
Now, M_{1}c^{2} = c^{2}(Z_{1}m_{H} + (A_{1}-Z)m_{n})-E_{1} etc and
Z_{1}+Z_{2} = Z_{3}+Z_{4} (Conservation of change)
A_{1} + A_{2} = A_{3} + A_{4} (Conservation of heavy particles)
Q = (Es + E4) – (Ex+E2)
13: Assuming that the splitting of a U^{236} nucleus liberates the energy of 200 MeV, find: (a) the energy liberated in the fission of one kilogram of U^{236} isotope, and the mass of coal with calorific value of 30 kJ/g which is equivalent to that for one kg of U^{235}; (b) the mass of U^{235} isotope split during the explosion of the atomic bomb with 30 kt trotyl equivalent if the calorific value of trotyl is 4.1 kJ/g.
Solution:
(a) the energy liberated in the fission of 1 kg of U^{235} is 100/235 x 6.023 x 10^{23} x 200MeV = 8.21 x 10^{10}kJ
The mass of coal with equivalent calorific value is
[30x10^{9}x4.1x10^{3}]/[200x1.602x10^{-13}x6.023x10^{23}] x 25/1000 kg = 1.49 kg
14: What amount of heat is liberated during the formation of one gram of He^{4} from deuterium H^{2}? What mass of coal with calorific value of 30 kJ/g is thermally equivalent to the magnitude obtained?
Solution:
The reaction is (in effect).
H^{2}+H^{2} ->He^{4}+Q
Then Q =2 ∆_{H}^{2} - ∆_{He}^{2} + Q
=0.02820-0.00260 = 0.02560 amu = 23.8 MeV
Hence,
the energy released in 1 gm of He^{4} is [6.023x10^{23}/4] x 23.8 x 16.02 x 10^{-13} Joule = 5.75 x 108 kJ
This energy can be derived from = [5.75x10^{8}/30000] kg = 1.9 x 10^{4} kg of Coal.
15: Taking the values of atomic masses from the tables, calculate the energy per nucleon which is liberated in the nuclear reaction Li^{6} + H^{2} → 2He^{4}. Compare the obtained magnitude with the energy per nucleon liberated in the fission of U^{235} nucleus.
Solution:
The energy released in the reaction
Li^{6} + H^{2} ->2He^{4} is ∆_{Li}^{6} + ∆_{H}^{2} -2 ∆_{He}^{4}
= 0.01513 + 0.01410 - 2 x 0.00 260 amu
= 0.02403 amu = 22.37 MeV
or 22.37/8 = 2.796 MeV/nucleon.
This should be compared with the value 200/235 = 0.85 MeV/nucleon
16: Find the energy of the reaction Li^{7} + p → 2He^{4} if the binding energies per nucleon in Li^{7} and He^{4} nuclei are known to be equal to 5.60 and 7.06 MeV respectively.
Solution:
The energy of reaction Li^{6} + p ->2He^{4} is 2 x B.E. of He^{4} - B.E. of Li^{7}
= 8εα - 7εLi = 8 x 7.06 - 7 x 5.60 = 17.3 MeV
17: Protons striking a stationary lithium target activate a reaction Li^{7} (p, n) Be^{7}. At what value of the proton's kinetic energy can the resulting neutron be stationary?
Solution:
We know, Q of this reaction (Li^{7}(p, n)Be^{7}).
If it is - 1.64 MeV. We have by conservation of momentum and energy P_{p} = P_{Be} (since initial Li and final neutron are both at rest)
P_{p}^{2}/2m_{p} = p^{2}_{Be}/2m_{Li} + 1.64
Then, P_{p}^{2}/2m_{p} (1 – m_{p}/m_{Be}) = 1.64
Hence, T_{p} = P_{p}^{2}/2m_{p} = 1.91 MeV
18: What kinetic energy must a proton possess to split a deuteron H^{2} whose binding energy is E_{b} = 2.2 MeV?
Solution:
Energy required to split a deuteron is T ≥ (1 + M_{p}/M_{d})E_{b} = 3.3 MeV
19: The irradiation of lithium and beryllium targets by a monoergic stream of protons reveals that the reaction Li^{7}(p, n)Be^{7} = 1.65 MeV is initiated whereas the reaction Be^{9} (p, n) B^{9} -1.85 MeV does not take place. Find the possible values of kinetic energy of the protons.
Solution:
Since the reaction Li^{7}(p,n)Be^{7} (Q = -1.65 MeV) is initiated, the incident proton energy must be
T ≥ (1 + M_{p}/M_{Li}) x 1.65 = 1.89 MeV
Since the reaction Be^{9}(p,n)B^{9} (Q = -1.85 MeV) is not initiated,
T ≤ (1 + M_{p}/M_{Li}) x 1.85 = 2.06 MeV
Thus 1.89 MeV ≤ T_{p} ≤ 2.06 MeV
20: To activate the reaction (n, a) with stationary B^{11} nuclei, neutrons must have the threshold kinetic energy T^{th} = 4.0 MeV. Find the energy of this reaction.
Solution:
We have (1 + M_{n}/M_{B}^{11})|Q|
Or Q = -11/12 x 4 MeV = -3.67 MeV