Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

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

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The document Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
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Q.249. An alpha-particle with kinetic energy Tα =7 .0 MeV is scattered elastically by an initially stationary Li6  nucleus. Find the kinetic energy of the recoil nucleus if the angle of divergence of the two particles is Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev = 60°. 

Ans. Initial momentum of the a particle is  Irodov Solutions: Nuclear Reactions- 1 Notes | EduRevis a unit vector in the incident direction). Final momenta are respectively Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev Conservation of momentum reads

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Squaring Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev(1)

where Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev is the angle between  Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Also by energy conservation Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

(m & M are respectively the masses of a particle and  Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev) So

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev(2)

Substracting (2) from (1) we see that

Thus if  Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Since pα, pLi are both positive number (being magnitudes of vectors) we must have

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

This being understood, we write

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Hence the recoil energy of the Li, nucleus is

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

As we pointed out above  Irodov Solutions: Nuclear Reactions- 1 Notes | EduRevrecoil energy of Li = 6 MeV

 

Q.250. A neutron collides elastically with an initially stationary deuteron. Find the fraction of the kinetic energy lost by the neutron (a) in a head-on collision; (b) in scattering at right angles. 

Ans. (a) In a head on collision

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Where pd and pn are the momenta of deuteron and neutron after the collision. Squaring

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

or since pd = 0 in a head on collisions

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Going back to energy conservation

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

So   Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

This is the energy lost by neutron. So the fraction of energy lost is

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

(b) In this case neutron is scattered by 90°. Then we have from the diagram

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Then by eneigy conservation

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

The energy lost by neutron in then

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

or fraction of energy lost is Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

 

 6.251. Find the greatest possible angle through which a deuteron is scattered as a result of elastic collision with an initially stationary proton.

Ans. From conservation of momentum 

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

From energy conservation 

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

(M = mass of denteron, m = mass of proton)

So Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Hence Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

For real roots Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

HenceIrodov Solutions: Nuclear Reactions- 1 Notes | EduRev

i.e.Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

For deuteron-proton scaltering  Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

 

 6.252. Assuming the radius of a nucleus to be equal to Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev pm, where A is its mass number, evaluate the density of nuclei and the number of nucleons per unit volume of the nucleus. 

Ans. This problem has a misprint Actually the radius R of a nucleus is given by

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

where  Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Then the number of nucleous per unit volume is Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev  Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

The corresponding mass density is (1.09 x 10-38 x mass of a nucleon) per cc = 1.82 x 1011kg/cc

 

6.253. Write missing symbols, denoted by x, in the following nuclear reactions: (a) B10 (x, α) Be8 

(b) O17  (d, n) x; (c) Na23  (p, x) Ne20; (d) x (p, n) Ar37

Ans.  (a) The particle x must carry two nucleons and a unit of positive charge.
The reaction is

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

(b) The particle x must contain a proton in addition to the constituents of O17. Thus the reaction is

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

(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

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

(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

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

 

 6.254. Demonstrate that the binding energy of a nucleus with mass number A and charge Z can be found from Eq. (6.6b). 

Ans. From the basic formula

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

We define

AH = mH - 1 amu

An = - 1 amu

A = M - A amu

Then clearly Eb - Z A# + (A - Z ) An - A

 

 6.255. 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 A127  nucleus. 

Ans. The mass number of the given nucleus must be

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Thus the nucleus is Be8. Then Ihe binding eneigy is

E b - 4 x 0-00867 + 4 + x 0-00783 - 0-00531 amu

= 0-06069 amu = 56-5 MeV

On using 1 amu = 931 MeV.

 

 6.256. Making use of the tables of atomic masses, find: (a) the mean binding energy per one nucleon in O16  nucleus; (b) the binding energy of a neutron and an alpha-particle in a B11  nucleus; (c) the energy required for separation of an O16 nucleus into four identical particles. 

Ans. (a) Total binding eneigy of Ihe O16 nucleus is

Eb = 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 B11 nucleus

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

(since on removing a neutron from B11 we get B10)

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

= 0.01231 amu = 11.46 MeV 

B.E. of ( an α-particle in B11)

= B.E. of B1 - B.E. of Li7 - B.E. of α 

(since on removing an a from B11 we get Li7 )

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

= - 0.00930 + 0.01601 + 0.00260 

= 0.00931 amu = 8.67 MeV

(c) This eneigy is

[B.E. of O16 + 4 (B.E. of a particles)]

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

= 4 x 0-00260 + 0.00509

= 0.01549 amu - 14.42 MeV

 

 6.257. Find the difference in binding energies of a neutron and a proton in a B11 nucleus. Explain why there is the difference.

Ans. B.E. o f a neutron in B11 - B.E. of a proton in B11

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev= 0.00867 - 0.00783

+ 0.01294 - 0.01354 = 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 Be10 which together constitute B11.

 

6.258. Find the energy required for separation of a Ne20  nucleus into two alpha-particles and a C12 nucleus if it is known that the binding energies per one nucleon in Ne20, He4, and C12  nuclei are equal to 8.03, 7.07, and 7.68 MeV respectively. 

Ans. Required energy is simply the difference in total binding energies

= B.E. of Ne20 - 2 (BE. of He4) - B.E. of C12 

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

(ε is binding energy per unit nucleon.) Substitution gives 11.88MeV .

 

 6.259. Calculate in atomic mass units the mass of (a) a Li8 atom whose nucleus has the binding energy 41.3 MeV; (b) a C10 nucleus whose binding energy per nucleon is equal to 6.04 MeV.

Ans.  We have for  Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

 41.3 MeV = 0.044361 amu = 3ΔH + 5Δn - Δ

Hence Δ = 3 x 0.00783 + 5 x 0.00867 - 0.09436 - 0.02248 amu 

(b) For C10                  10 x 6.04 = 60.4 MeV

- 0-06488 amu 

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Hence Δ = 6 x 0.00783 + 4 x 0-00867 - 0.06488 = 0.01678 amu 

Hence the mass of C10 is 10.01678 amu

 

6.260. The nuclei involved in the nuclear reaction A+ A2 → → A3 + Ahave the binding energies E1, E2, E3, and E4. Find the energy of this reaction.

Ans. Suppose M1 , M2 , M3 , Mare the rest masses of the nuclei Alf A2 , A3 and A4 participating in the reaction

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Here Q is the energy released. Then by conservation of energy.

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

NowIrodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Z+ Z2 = Z3 + Z4 (conservation of change)

A+ A2 = A3 + A4 (conservation of heavy particles)

Hence                     Q = (Es + E4) - (Ex + E2)

 

6.261. Assuming that the splitting of a U236  nucleus liberates the energy of 200 MeV, find: (a) the energy liberated in the fission of one kilogram of U236 isotope, and the mass of coal with calorific value of 30 kJ/g which is equivalent to that for one kg of U235; (b) the mass of U235 i sotope split during the explosion of the atomic bomb with 30 kt trotyl equivalent if the calorific value of trotyl is 4.1 kJ/g. 

Ans.  (a) the energy liberated in the fission of 1 kg of U235 is

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev 6.023 x 1023 x 200MeV = 8.21 x 1010kJ

The mass of coal with equivalent calorific value is

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev= 2.74 x 106 kg

(b) The required mass' is

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

 

6.262. What amount of heat is liberated during the formation of one gram of He4  from deuterium H2? What mass of coal with calorific value of 30 kJ/g is thermally equivalent to the magnitude obtained?

Ans. The reaction is (in effect).

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

 Then 

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

= 0.02820-0.00260 
= 0.02560 amu = 23.8 MeV 

Hence the energy released in 1 gm of He4 is

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev 23.8 x 16.02 x 10-13 Joule = 5.75 x 108 kJ

This eneigy can be derived from

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev=1.9 x 104 kg of Coal.

 

6.263. Taking the values of atomic masses from the tables, calculate the energy per nucleon which is liberated in the nuclear reaction Li+ H2 → 2He4. Compare the obtained magnitude with the energy per nucleon liberated in the fission of U235  nucleus.

Ans. The energy released in the reaction

is Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

= 0.01513 + 0.01410 - 2 x 0.00 260 amu 

= 0.02403 amu = 22.37 MeV

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev2.796 MeV/nucleon.

This should be compared with the value Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev= 0.85 MeV/nucleon

 

6.264. Find the energy of the reaction Li+ p → 2Heif the binding energies per nucleon in Li7 and He4 nuclei are known to be equal to 5.60 and 7.06 MeV respectively. 

Ans.  The eneigy of reaction

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

is,            2 x B.E. of He4 - B.E. of Li

= 8εα - 7εLi = 8 x 7.06 - 7 x 5.60 = 17.3 MeV

 

6.265. Find the energy of the reaction N14  (α, p) O17 if the kinetic energy of the incoming alpha-particle is Tα = 4.0 MeV and the proton outgoing at an angle θ = 60° to the motion direction of the alpha-particle has a kinetic energy Tp = 2.09 MeV. 

Ans. The reaction is N14(a, p)O17

It is given that (in the Lab frame where N14 is at rest) Tα = 4.0MeV. The momentum of incident α particle is 

The momentum of outgoing proton is

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Where Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

and mis the mass of O17

The momentum of O17 is

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

By energy conservation (conservation of energy including rest mass energy and kinetic energy) 

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Hence by definition of the Q of reaction

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

 

6.266. Making use of the tables of atomic masses, determine the energies of the following reactions: (a) Li(p, n) Be7; (b) Be9 (n,γ) Be10; (c) Li (α, n) B10); (d) O16  (d, α) N14.

Ans. (a) The reaction is Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev and the energy of reaction is

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

=-1.64 MeV

(b) The reaction is Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Mass of γ is taken zero. Then

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

= (0.01219 + 0.00867 - 0.01354) Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

= 6.81 MeV

(c ) The reaction is  Li(α,n) B10. The energy is

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

= (0.01601 + 0.00260 - 0.00867 - .01294) amu x c2 

= - 2.79 MeV

(d) The reaction is O16 (d , α) N14 The energy of reaction is

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

= (- 0.00509 + 0.01410 - 0.00260 - 0.00307) amu x c2

= 3.11 MeV

 

6.267. Making use of the tables of atomic masses, find the velocity with which the products of the reaction B10 (n, α) Li7 come apart; the reaction proceeds via interaction of very slow neutrons with stationary boron nuclei.

Ans. The reaction is B10 (n, ct) Li1. The energy of the reaction is

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

= (0.01294 + 0-00867 - 0.00260 - 0.01601) amu x c2

= 2-79 MeV

Since the incident neutron is very slow and B10 is stationary, the final total momentum must also be zero. So the reaction products must emerge in opposite directions. If their speeds are, repectively, va and vLi

then 4va = 7vLi

and Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev= 2.79 x 1.602 x 10-6

So Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev 2.70 x 1018 cm2/s2 

or         va = 9.27x10m/s

Then        vLi = 5.3 x 10m/s

 

6.268. Protons striking a stationary lithium target activate a reaction Li7 (p, n) Be7. At what value of the proton's kinetic energy can the resulting neutron be stationary?

Ans. Q of this reaction (Li7(p, n)Be7) was calculated in problem 266 (a). If is - 1.64 MeV.
We have by conservation of momentum and energy Pp = PBe (since initial Li and final neutron are both at rest)

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Then  Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

 Hence   Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev= 1.91MeV 

 

6.269. An alpha particle with kinetic energy T = 5.3 MeV initiates a nuclear reaction Be9 (α, n) C12  with energy yield Q =+5.7 MeV. Find the kinetic energy of the neutron outgoing at right angles to the motion direction of the alpha-particle. 

Ans. It is understood that Be9 is initially at rest. The moment of the outgoing neutron is 

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRevThe momentum of C12 is

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Then by energy conservation

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

 

6.270. Protons with kinetic energy T =1.0 MeV striking a lithium target induce a nuclear reaction p + Li7 → 2He4. Find the kinetic energy of each alpha-particle and the angle of their divergence provided their motion directions are symmetrical with respect to that of incoming protons. 

Ans. The Q value of the reaction  Li(p, α) Heis

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

= (0.01601 + 0.00783 - 0.00520) amu x c2

= 0.01864 amu x c2 - 17.35 MeV

Since the direction of He4 nuclei is symmetrical, their momenta must also be equal. Let T be the K.E. of each He4. Then

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

(pp is the momentum of proton). Also

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Hence Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Hence  Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

Substitution gives            θ = 170.53°

Also Irodov Solutions: Nuclear Reactions- 1 Notes | EduRev

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