I.E. Irodov Solutions on Nuclear Reactions

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.

I.E. Irodov Solutions on Nuclear Reactions

1: 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 ϕ = 60°.

Solution:

1. Initial momentum of a particle is √(2mTα) i

Where i is a unit vector in the incident direction

Final momenta are respectively $\vec{P_{\alpha}} \ and \ \vec{P_{Li}}$. Conservation of momentum reads

$\vec{P_{\alpha}} \ and \ \vec{P_{Li}} = \sqrt{2mT_\alpha}\hat{i}$

Squaring P2α + P2Li + 2Pα PLi cos ϕ = 2m Tα ……(1)

Also, by energy conservation, P2α /2m + P2Li/2M = Tα

P2α + m/M P2Li = 2mTα …..(2)

Subtracting (2) from (1)

PLi[(1-m/M)PLi + 2P α cos ϕ] = 0

If PLi = 0, then Pα = (-1/2)(1 – m/M)PLi sec ϕ

Since Pα and PLi are both positive number (being magnitudes of vectors)

we must have, -1 ≤ cos ϕ < 0 if m < M.

This being understood, we write

(P2Li/2M) [1 + M/4m(1-m/M)2 sec2 ϕ] = Tα

(P2Li/2M) = Tα/[1 + M/4m(1-m/M)2 sec2 ϕ]

As ϕ is given. If we take ϕ = 1200, 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:

1. (a) In a head on collision

√(2mT) = pd + pn

T = p2d/2M + p2n/2m

Where pd and pn are the momenta of deuteron and neutron after the collision. Squaring ,2mT = p2d + p2n + 2pdpn

p2n + (m/M)p2d = 2mT

or since pd = 0 in a head on collisions

pn=(-1/2)(1-m/M)pd

Going back to energy conservation

(p2d/2M) (1+ M/4m (1 – m/M)2) = T

So, pd2/2M = [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 90o. Then we have

$\vec{p_d} = p_n \hat{j} + \sqrt{2mT} \hat{i}$

Then by energy conservation:

$\frac{p_n^2 + 2mT}{2M} + \frac{p_n^2}{2m}=T$

Or (pn2/2m)(1 + m/M) = T(1 - m/M)

Or T - pn2/2m) = 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:

1. From conservation of momentum:

$\sqrt{2MT}\hat{i}=\vec{p_d} + \vec{p_p}$

or pp2 = 2MT + pd2 - 2√(2MT) pd cosθ

From energy conservation

T = pp2 /2m + pd2/2M

(M = mass of denteron and m = mass of proton)
so, pp2 = 2mT – (m/M) pd2

Therefore, pd2(1 + m/M) - 2 √(2MT) pd cos θ + 2(M-m)T = 0

For real roots, 4(2MT)cos2 θ - 4x2(M-m)T(1 + m/M) ≥0

Cos2 θ ≥ (1 -m2/M2)

Hence, sin2 θ ≤ m2/M2

i.e. θ ≤ sin-1(m/M)

For deuteron-proton scaltering, θmax = 30o.

4: Assuming the radius of a nucleus to be equal to 0.13 A1/3 pm, where A is its mass number, evaluate the density of nuclei and the number of nucleons per unit volume of the nucleus.

Solution:

1. This problem has a misprint, actually the radius R of a nucleus is given as 1.3 A1/3 pm where fm = 10-15 m

Then the number of nucleous per unit volume is

A/[(4π/3) R3] = 1.09 x 1038 per cc

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

5: 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.

Solution:

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

The reaction is B10(d, α)Beβ

(b) The particle x must contain a proton in addition to the constituents of O17. Thus the reaction is O17(d, n)F18

(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 Na23(p, α)Ne20.

(d) The particle x must carry mass number 37 and have one unit less of positive charge. Thus x = Cl37 and the reaction is Cl37(p,n) Ar37.

6: Demonstrate that the binding energy of a nucleus with mass number A and charge Z can be found from Eb = Z ∆H + (A - Z) ∆n - ∆.

Solution:

1. From the basic formula, Eb = Z ∆H + (A - Z) ∆n - ∆

We define AH = mH - 1 amu

An = - 1 amu

A = M - A amu

Then clearly Eb - Z A# + (A - Z ) An – 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 Al27 nucleus.

Solution:

1. The mass number of the given nucleus must be 27/(3/2)3 = 8

Thus the nucleus is Be8. Then The binding energy is Eb - 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 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.

Solution:

1. (a) Total binding energy of the 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 = B.E. of B11 – B.E. of B10

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

=∆n - ∆B11 + ∆B10 = 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 )

= -∆B11 + ∆Li + ∆α

= - 0.00930 + 0.01601 + 0.00260 = 0.00931 amu = 8.67 MeV

(c) This energy is

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

= -∆016 + 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 B11 nucleus. Explain why there is the difference.

Solution:

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

=(∆n - ∆B11 + ∆B10) – (∆p - ∆B11 + ∆Be10)

= 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.

10: 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.

Solution:

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

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

= 20 εNe - 8 εα – 12 εC

(ε is binding energy per unit nucleon.)

Substitution gives 11.88MeV.

11: 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.

Solution:

1. We have for Li8

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

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

Hence the mass of C10 is 10.01678 amu.

12: The nuclei involved in the nuclear reaction A1 + A2 → A3 + A4 have the binding energies E1, E2, E3, and E4. Find the energy of this reaction.

Solution:

1. Suppose M1, M2, M3, M4 are the rest masses of the nuclei A1. lf A2 , A3 and A4 participating in the reaction

A1 + A2 → A3 + A4 + Q

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

Q = c2(M1 + M2 - M3 - M4)

Now, M1c2 = c2(Z1mH + (A1-Z)mn)-E1 etc and

Z1+Z2 = Z3+Z4 (Conservation of change)

A1 + A2 = A3 + A4 (Conservation of heavy particles)

Q = (Es + E4) – (Ex+E2)

13: 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 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:

1. (a) the energy liberated in the fission of 1 kg of U235 is 100/235 x 6.023 x 1023 x 200MeV = 8.21 x 1010kJ

The mass of coal with equivalent calorific value is

[30x109x4.1x103]/[200x1.602x10-13x6.023x1023] x 25/1000 kg = 1.49 kg

14: 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?

Solution:

1. The reaction is (in effect).

H2+H2 ->He4+Q

Then Q =2 ∆H2 - ∆He2 + Q

=0.02820-0.00260 = 0.02560 amu = 23.8 MeV

Hence,

the energy released in 1 gm of He4 is [6.023x1023/4] x 23.8 x 16.02 x 10-13 Joule = 5.75 x 108 kJ

This energy can be derived from = [5.75x108/30000] kg = 1.9 x 104 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 Li6 + H2 → 2He4. Compare the obtained magnitude with the energy per nucleon liberated in the fission of U235 nucleus.

Solution:

1. The energy released in the reaction

Li6 + H2 ->2He4 is ∆Li6 + ∆H2 -2 ∆He4

= 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 Li7 + p → 2He4 if the binding energies per nucleon in Li7 and He4 nuclei are known to be equal to 5.60 and 7.06 MeV respectively.

Solution:

1. The energy of reaction Li6 + p ->2He4 is 2 x B.E. of He4 - B.E. of Li7

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

17: 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?

Solution:

1. We know, Q of this reaction (Li7(p, n)Be7).

If it is - 1.64 MeV. We have by conservation of momentum and energy Pp = PBe (since initial Li and final neutron are both at rest)

Pp2/2mp = p2Be/2mLi + 1.64

Then, Pp2/2mp (1 – mp/mBe) = 1.64

Hence, Tp = Pp2/2mp = 1.91 MeV

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

Solution:

1. Energy required to split a deuteron is T ≥ (1 + Mp/Md)Eb = 3.3 MeV

19: 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.

Solution:

1. Since the reaction Li7(p,n)Be7 (Q = -1.65 MeV) is initiated, the incident proton energy must be

T ≥ (1 + Mp/MLi) x 1.65 = 1.89 MeV

Since the reaction Be9(p,n)B9 (Q = -1.85 MeV) is not initiated,

T ≤ (1 + Mp/MLi) x 1.85 = 2.06 MeV

Thus 1.89 MeV ≤ Tp ≤ 2.06 MeV

20: 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.

Solution:

1. We have (1 + Mn/MB11)|Q|

Or Q = -11/12 x 4 MeV = -3.67 MeV