I.E. Irodov Solutions on Nuclear Reactions

1: An alpha-particle with kinetic energy T_α = 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°.

Initial momentum of a particle is √(2mT_α) i

Where i is a unit vector in the incident direction

Final momenta are respectively

Squaring P²_α + P²_Li + 2P_α P_Li cos ϕ = 2m T_α ……(1)

Also, by energy conservation, P²_α /2m + P²_Li/2M = T_α

P²_α + m/M P²_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

(P²_Li/2M) [1 + M/4m(1-m/M)² sec² ϕ] = T_α

(P²_Li/2M) = T_α/[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.

(a) In a head on collision

√(2mT) = p_d + p_n

T = p²_d/2M + p²_n/2m

Where p_d and p_n are the momenta of deuteron and neutron after the collision. Squaring ,2mT = p²_d + p²_n + 2p_dp_n

p²_n + (m/M)p²_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

(p²_d/2M) (1+ M/4m (1 – m/M)²) = T

So, p_d²/2M = [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_n²/2m)(1 + m/M) = T(1 – m/M)

Or T – p_n²/2m) = 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.

From conservation of momentum:

or p_p² = 2MT + p_d² – 2√(2MT) p_d cosθ

From energy conservation

T = p_p² /2m + p_d²/2M

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

so, p_p² = 2mT – (m/M) p_d²

Therefore, p_d²(1 + m/M) – 2 √(2MT) p_d cos θ + 2(M-m)T = 0

For real roots, 4(2MT)cos² θ – 4×2(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, 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.

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) 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.

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

The reaction is B¹⁰(d, α)Be^β

(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 E_b = Z ∆_H + (A – Z) ∆_n – ∆.

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 Al²⁷ nucleus.

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

Thus the nucleus is Be⁸. 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 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.

(a) Total binding energy of the O¹⁶ 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 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.

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.

Required energy is simply the difference in total binding energies

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

= 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 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.

We have for Li⁸

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 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.

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(Z₁m_H + (A₁-Z)m_n)-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.

(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

[30×10⁹x4.1×10³]/[200×1.602×10^-13x6.023×10²³] 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?

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 [6.023×10²³/4] x 23.8 x 16.02 x 10^-13 Joule = 5.75 x 108 kJ

This energy can be derived from = [5.75×10⁸/30000] 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.

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.

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?

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

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_p²/2m_p = p²_Be/2m_Li + 1.64

Then, P_p²/2m_p (1 – m_p/m_Be) = 1.64

Hence, T_p = P_p²/2m_p = 1.91 MeV

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

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 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.

Since the reaction Li⁷(p,n)Be⁷ (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 Be⁹(p,n)B⁹ (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 B¹¹ nuclei, neutrons must have the threshold kinetic energy T^th = 4.0 MeV. Find the energy of this reaction.

We have (1 + M_n/M_B¹¹)|Q|

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