6.271. A particle of mass m strikes a stationary nucleus of mass M and activates an endoergic reaction. Demonstrate that the threshold (minimal) kinetic energy required to initiate this reaction is defined by Eq. (6.6d).
Ans. Energy required is minimum when the reaction products all move in the direction of the incident particle with the same velocity (so that the combination is at rest in the centre of mass frame). We then have
(Total mass is constant in the nonrelativistic limit).
or
Hence
6.272. What kinetic energy must a proton possess to split a deuteron H^{2} whose binding energy is E_{b} = 2.2 MeV?
Ans. The result of the previous problem applies and we End that energy required to split a deuteron is
6.273. 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.
Ans. Since the reaction is initiated, the incident proton energy must be
since the reaction is not initiated,
= 2.06 MeV Thus 1.89 MeV ≤ T_{p }≤ 2.06 MeV
6.274. 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.
Ans. We have
or 3.67 MeV
6.275. Calculate the threshold kinetic energies of protons required to activate the reactions (p, n) and (p, d) with Li^{7} nuclei.
Ans. The Q o f the reaction Li^{7} (p, n ) Be^{7} was calculated in problem 266 (a). It is  1.64 MeV Hence, the threshold K.E. of protons for initiating this reaction is
For the reaction Li^{7} (p, d) Li^{6}
we find
= (0.01601 + 0.00783  0.01410  0.01513) amu x c^{2}
=  5 02 MeV
The threshold proton energy for initiating this reaction is
5.73 MeV
6.276. Using the tabular values of atomic masses, find the threshold kinetic energy of an alpha particle required to activate the nuclear reaction Li^{7} (α, n) Be^{9}. What is the velocity of the B^{10} nucleus in this case?
Ans. The Q of Li^{7} (a, n) B^{10} was calculated in problem 266 (c). It is Q = 2.79 MeV Then the threshold energy of aparticle is
4.38MeV
The velocity of B^{10} in this case is simply the volocity of centre of mass
This is because both B^{10 }and n are at rest in the CM frame at theshold. Substituting the values of various quantities
we get v = 5.27 x 10^{6} m/s
6.277. A neutron with kinetic energy T = 10 MeV activates a nuclear reaction C^{12 } (n, α) Be^{9} whose threshold is T_{th} = 6.17 MeV. Find the kinetic energy of the alphaparticles outgoing at right angles to the incoming neutrons' direction.
Ans. The momentum of incident neutron is that of α particle is and of
By conservation of energy
(M is the mass of Be^{9}). Thus
Using
we get
M’ is the mass of C^{12} nucleus.
or = 2.21MeV
6.278. How much, in per cent, does the threshold energy of gamma quantum exceed the binding energy of a deuteron (E_{b} = 2.2 MeV) in the reaction Y + H^{2} → n + p?
Ans. The formula of problem 6.271 does not apply here because the photon is always reletivistic. At threshold, the energy o f the photon E_{y} implies a momentum The velocity o f centre of mass with respect to the rest frame of initial H^{2} is
Since both n & p are at rest in CM frame at threshold, we write
by conservation of energy. Since the first term is a small correction, we have
Thus
or nearly 0.06%
6.279. A proton with kinetic energy T = 1.5 MeV is captured by a deuteron H^{2}. Find the excitation energy of the formed nucleus.
Ans. The reaction is
Excitation energy of He^{3} is just the energy available in centre of mass. The velocity of the centre of mass is
In the CM frame, the kinetic energy available is (m_{d ≈ }2 m_{p})
The total energy available is then
where
= c2 x (0.00783 + 0.01410  0.01603) amu
= 5.49 McV
Finally E = 6.49 MeV.
6.280. The yield of the nuclear reaction C^{13}(d, n)N^{14} has maximum magnitudes at the following values of kinetic energy T_{1 }of bombarding deuterons: 0.60, 0.90, 1.55, and 1.80 MeV. Making use of the table of atomic masses, find the corresponding energy levels of the transitional nucleus through which this reaction proceeds.
Ans. The reaction is
Maxima of yields deteimine the eneigy levels of N^{15 }*. As in the previous problem the excitation eneigy is
where E_{K} = available kinetic eneigy. This is found is as in the previous problem. The velocity of the centre of mass is
so
Q is the Q value for the ground state of N^{15} : We have
= c^{2} x (0.01410 + 0.00335  0.00011) amu
= 16.14 MeV
The excitation energies then are
16.66 MeV, 16.92 MeV
17.49 MeV and 17.70 MeV.
6.281. A narrow beam of thermal neutrons is attenuated η = 360 times after passing through a cadmium plate of thickness d = 0.50 mm. Determine the effective crosssection of interaction of these neutrons with cadmium nuclei.
Ans. We have the relation
Here attenuation factor
n = no. of Cd nuclei per unit volume
σ = effective cross section
d = thickness of the plate
Now
(p = density, M = Molar weight of Cd, N_{A} = Avogadro number.)
Thus
6.282. Determine how many times the intensity of a narrow beam of thermal neutrons will decrease after passing through the heavy water layer of thickness d = 5.0 cm. The effective crosssections of interaction of deuterium and oxygen nuclei with thermal neutrons are equal to σ_{1} = 7.0 b and σ_{2}  4.2 b respectively.
Ans. Here
where 1 refers to O^{1} and 2 to D nuclei
Using n_{2} = 2n, n_{1} = n = concentration of O nuclei in heavy water we get
Now using the data for heavy water
3.313 x 10^{22} per cc
Thus substituting the values
6.283. A narrow beam of thermal neutrons passes through a plate of iron whose absorption and scattering effective crosssections are equal to σ_{a} = 2.5b and σ_{8} = 11b respectively. Find the fraction of neutrons quitting the beam due to scattering if the thickness of the plate is d = 0.50 cm.
Ans. In traversing a distance d the fraction which is either scattered or absorbed is clearly
by the usual definition of the attenuation factor. Of this, the fraction scattered is (by definition of scattering and absorption cross section)
In iron
Substitution gives w = 0.352
6.284. The yield of a nuclear reaction producing radionuclides may be described in two ways: either by the ratio w of the number of nuclear reactions to the number of bombarding particles, or by the quantity k, the ratio of the activity of the formed radionuclide to the number of bombarding particles, Find: (a) the halflife of the formed radionuclide, assuming w and k to be known; (b) the yield w of the reaction Li^{7}(p, n)Be^{7} if after irradiation of a lithium target by a beam of protons (over t = 2.0 hours and with beam current I = 10μA) the activity of Be^{7} became equal to A = = 1.35.10^{8} dis/s and its halflife to T = 53 days.
Ans. (a) Assuming of course, that each reaction produces a radio nuclide of the same type, the decay constant α of the radionuclide is k/w . Hence
(b) number of bombarding particles is
(e = charge on proton). Then the number of Be^{7} produced is
If λ = decay constant of , then the activity is
Hence
6.285. Thermal neutrons fall normally on the surface of a thin gold foil consisting of stable Au^{197} nuclide. The neutron flux density is J = 1.0.10^{10 } part./(s cm^{2}). The mass of the foil is m = 10 mg. The neutron capture produces betaactive Au^{188 } nuclei with halflife T = 2.7 days. The effective capture crosssection is σ = 98 b. Find: (a) the irradiation time after which the number of Au^{187} nuclei decreases by = 1.0%; (b) the maximum number of Au^{198} nuclei that can be formed during protracted irradiation.
Ans. (a) Suppose  N_{o}. of Au^{197} nuclei in the foil. Then the number of Au^{197} nuclei transformed in time t is
For this to equal ηN_{0}, we must have
= 323 years
(b) Rate of formation of the Au^{198} nuclei is N_{0} • J • a per sec
and rate of decay is λn, where n is the number of Au^{198} at any in stant
Thus
The maximum number of Au^{198} is clearly
because if n is smaller, and n increase further and if n is larger
and n will decrease. (Actually n_{max} is approached steadily as )
Substitution give susing
6.286. A thin foil of certain stable isotope is irradiated by thermal neutrons falling normally on its surface. Due to the capture of neutrons a radionuclide with decay constant λ appears. Find the law describing accumulation of that radionuclide N (t) per unit area of the foil's surface. The neutron flux density is J, the number of nuclei per unit area of the foil's surface is n, and the effective crosssection of formation of active nuclei is α.
Ans. Rate of formation of the radionuclide is n.J.σ per unit area persec. Rate of decay is λN.
Thus
per unit area per second
Then or
Hence
The number of radionuclide at t = 0 when the process starts is zero. So constant
Then
6.287. A gold foil of mass m = 0.20g was irradiated during t = 6.0 hours by a thermal neutron flux falling normally on its surface. Following ζ = 12 hours after the completion of irradiation the activity of the foil became equal to A = 1.9.10^{7} dis/s. Find the neutron flux density if the effective crosssection of formation of a radioactive nucleus is σ = 96b, and the halflife is equal to T = 2.7 days.
Ans. We apply the formula of the previous problem except that have N = no. of radio nuclide and no. of host nuclei originally.
Here 6.115 x10^{20}
Then after time t
T = half life of the radionuclide.
After the source of neutrons is cut off the activity after time T will be
Thus
6.288. How many neutrons are there in the hundredth generation if the fission process starts with N_{o } = 1000 neutrons and takes place in a medium with multiplication constant k = 1.05?
Ans. No. of nuclei in the first generation = No .of nuclei initially = N_{0}
N_{0} in the second generation = N_{0 }x multiplication factor = N_{0}.k
N_{0} in the the 3^{rd }generation
N_{0} in the nth generation
Substitution gives 1.25 x 10^{5} neutrons
6.289. Find the number of neutrons generated per unit time in a uranium reactor whose thermal power is P = 100 MW if the average number of neutrons liberated in each nuclear splitting is v = 2.5. Each splitting is assumed to release an energy E = = 200 MeV.
Ans. N_{o} of fissions per unit time is clearly P/E. Hence no. of neutrons produced per unit time to
Substitution gives 7.80 x 10^{18} neutrons/sec
6.290. In a thermal reactor the mean lifetime of one generation of thermal neutrons is ζ = 0.10 s. Assuming the multiplication constant to be equal to k = 1.010, find: (a) how many times the number of neutrons in the reactor, and consequently its power, will increase over t = 1.0 min;
(b) the period T of the reactor, i.e. the time period over which its power increases efold.
Ans. (a) This number is k^{n  1} where n = no. of generations in time t = t/T
Substitution gives 388.
(b) We write
or and
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