Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

I. E. Irodov Solutions for Physics Class 11 & Class 12

JEE : Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

The document Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
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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

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

(Total mass is constant in the nonrelativistic limit).

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

or Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

HenceIrodov Solutions: Nuclear Reactions- 2 Notes | EduRev

 

6.272. What kinetic energy must a proton possess to split a deuteron H2  whose binding energy is Eb = 2.2 MeV? 

Ans. The result of the previous problem applies and we End that energy required to split a deuteron is

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

 

6.273. The irradiation of lithium and beryllium targets by a monoergic stream of protons reveals that the reaction Li7(p, n)Be7  -- 1.65 MeV is initiated whereas the reaction Be9(p, n)B9 —  1.85 MeV does not take place. Find the possible values of kinetic energy of the protons.

Ans. Since the reaction Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev is initiated, the incident proton energy must be

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

since the reaction Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev is not initiated,

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev= 2.06 MeV Thus 1.89 MeV ≤ T≤ 2.06 MeV

 

6.274. To activate the reaction (n, a) with stationary B11  nuclei, neutrons must have the threshold kinetic energy Tth = 4.0 MeV. Find the energy of this reaction.

Ans.  We have   Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

or Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev -3.67 MeV

 

6.275. Calculate the threshold kinetic energies of protons required to activate the reactions (p, n) and (p, d) with Li7  nuclei. 

Ans. The Q o f the reaction Li7 (p, n ) Be7 was calculated in problem 266 (a). It is - 1.64 MeV Hence, the threshold K.E. of protons for initiating this reaction is

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

For the reaction Li7 (p, d) Li6

we find Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

= (0.01601 + 0.00783 - 0.01410 - 0.01513) amu x c2

= - 5 02 MeV

The threshold proton energy for initiating this reaction is

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev5.73 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 Li7 (α, n) Be9. What is the velocity of the B10 nucleus in this case? 

Ans. The Q of Li7 (a, n) B10 was calculated in problem 266 (c). It is Q = 2.79 MeV Then the threshold energy of a-particle is

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev 4.38MeV

The velocity of B10 in this case is simply the volocity of centre of mass

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRevIrodov Solutions: Nuclear Reactions- 2 Notes | EduRev

This is because both B10 and n are at rest in the CM frame at theshold. Substituting the values of various quantities

we get               v =  5.27 x 106 m/s

 

6.277. A neutron with kinetic energy T = 10 MeV activates a nuclear reaction C12  (n, α) Be9  whose threshold is Tth = 6.17 MeV. Find the kinetic energy of the alpha-particles outgoing at right angles to the incoming neutrons' direction. 

Ans. The momentum of incident neutron is  Irodov Solutions: Nuclear Reactions- 2 Notes | EduRevthat of α particle is  Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev and of

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

By conservation of energy 

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

(M is the mass of Be9). Thus

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

Using  Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

we get Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

M’ is the mass of C12 nucleus.

orIrodov Solutions: Nuclear Reactions- 2 Notes | EduRev = 2.21MeV

 

6.278. How much, in per cent, does the threshold energy of gamma quantum exceed the binding energy of a deuteron (Eb = 2.2 MeV) in the reaction Y  + H2  → 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 Ey implies a momentum  Irodov Solutions: Nuclear Reactions- 2 Notes | EduRevThe velocity o f centre of mass with respect to the rest frame of initial H2 is

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

Since both n & p are at rest in CM frame at threshold, we write

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

by conservation of energy. Since the first term is a small correction, we have

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

ThusIrodov Solutions: Nuclear Reactions- 2 Notes | EduRev  Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

or nearly 0.06%

 

6.279. A proton with kinetic energy T = 1.5 MeV is captured by a deuteron H2. Find the excitation energy of the formed nucleus. 

Ans. The reaction is

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

Excitation energy of He3 is just the energy available in centre of mass. The velocity of the centre of mass is

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

In the CM frame, the kinetic energy available is (md ≈ 2 mp)

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

The total energy available is then Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

where  Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

= 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 C13(d, n)N14 has maximum magnitudes at the following values of kinetic energy Tof 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

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

Maxima of yields deteimine the eneigy levels of N15 *. As in the previous problem the excitation eneigy is

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

where EK = available kinetic eneigy. This is found is as in the previous problem. The velocity of the centre of mass is

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

so   Irodov Solutions: Nuclear Reactions- 2 Notes | EduRevIrodov Solutions: Nuclear Reactions- 2 Notes | EduRev

Q is the Q value for the ground state of N15 : We have

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

= c2 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 cross-section of interaction of these neutrons with cadmium nuclei.

Ans. We have the relation

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

Here Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev attenuation factor

n = no. of Cd nuclei per unit volume

σ = effective cross section

d = thickness of the plate

Now  Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

(p = density, M = Molar weight of Cd, NA = Avogadro number.)

Thus  Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

 

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 cross-sections of interaction of deuterium and oxygen nuclei with thermal neutrons are equal to σ1 = 7.0 b and σ2  --- 4.2 b respectively. 

Ans. Here

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

where 1 refers to O1 and 2 to D nuclei

Using n2 = 2n, n1 = n = concentration of O nuclei in heavy water we get

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

Now using the data for heavy water

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev 3.313 x 1022 per cc

Thus substituting the values

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

 

6.283. A narrow beam of thermal neutrons passes through a plate of iron whose absorption and scattering effective cross-sections are equal to σa = 2.5b and σ8 = 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

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

by the usual definition of the attenuation factor. Of this, the fraction scattered is (by definition of scattering and absorption cross section)

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

In iron  Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

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 half-life of the formed radionuclide, assuming w and k to be known; (b) the yield w of the reaction Li7(p, n)Be7  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 Be7 became equal to A = = 1.35.108  dis/s and its half-life 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

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

(b) number of bombarding particles is Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

(e = charge on proton). Then the number of Be7 produced is Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

If λ = decay constant of Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev , then the activity is  Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

Hence Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

 

6.285. Thermal neutrons fall normally on the surface of a thin gold foil consisting of stable Au197  nuclide. The neutron flux density is J = 1.0.1010  part./(s- cm2). The mass of the foil is m = 10 mg. The neutron capture produces beta-active Au188  nuclei with half-life T = 2.7 days. The effective capture cross-section is σ = 98 b. Find: (a) the irradiation time after which the number of Au187  nuclei decreases by = 1.0%; (b) the maximum number of Au198 nuclei that can be formed during protracted irradiation.

Ans. (a) Suppose - No. of Au197 nuclei in the foil. Then the number of Au197 nuclei transformed in time t is

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

For this to equal ηN0, we must have

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev = 323 years

(b) Rate of formation of the Au198 nuclei is N0 • J • a per sec
and rate of decay is λn, where n is the number of Au198 at any in stant

Thus

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

The maximum number of Au198 is clearly

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

because if n is smaller, Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev and n increase further and if n is larger

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev and n will decrease. (Actually nmax is approached steadily as Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev)

Substitution give susing Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

 

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 per-sec. Rate of decay is λN.
Thus

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev per unit area per second

Then  Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev or  Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

Hence Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

The number of radionuclide at t = 0 when the process starts is zero. So constant Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

ThenIrodov Solutions: Nuclear Reactions- 2 Notes | EduRev

 

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 = 1.9.107 dis/s. Find the neutron flux density if the effective cross-section of formation of a radioactive nucleus is σ = 96b, and the half-life 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 Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev 6.115 x1020

Then after time t Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

T = half life of the radionuclide.

After the source of neutrons is cut off the activity after time T will be

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRevIrodov Solutions: Nuclear Reactions- 2 Notes | EduRev

Thus Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev  Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

 

6.288. How many neutrons are there in the hundredth generation if the fission process starts with N = 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 = N0

N0 in the second generation = Nx multiplication factor = N0.k

N0 in the the 3rd generationIrodov Solutions: Nuclear Reactions- 2 Notes | EduRev

N0 in the nth generation Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

Substitution gives 1.25 x 105 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. No- of fissions per unit time is clearly P/E. Hence no. of neutrons produced per unit time to

Irodov Solutions: Nuclear Reactions- 2 Notes | EduRevSubstitution gives 7.80 x 1018 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 e-fold. 

 Ans. (a) This number is kn - 1 where n = no. of generations in time t = t/T

Substitution gives 388.

(b) We writeIrodov Solutions: Nuclear Reactions- 2 Notes | EduRev

orIrodov Solutions: Nuclear Reactions- 2 Notes | EduRev and  Irodov Solutions: Nuclear Reactions- 2 Notes | EduRev

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