Irodov Solutions: Nuclear Reactions- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE PDF Download

Q.249. 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  = 60°. 

Ans. Initial momentum of the a particle is  is a unit vector in the incident direction). Final momenta are respectively  Conservation of momentum reads

Squaring (1)

where  is the angle between  

Also by energy conservation 

(m & M are respectively the masses of a particle and  ) So

(2)

Substracting (2) from (1) we see that

Thus if  

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

This being understood, we write

Hence the recoil energy of the L_i, nucleus is

As we pointed out above  recoil 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

Where p_d and p_n are the momenta of deuteron and neutron after the collision. Squaring

or since p_d = 0 in a head on collisions

Going back to energy conservation

So  

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

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

Then by eneigy conservation

The energy lost by neutron in then

or fraction of energy lost is

 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 

From energy conservation 

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

So 

Hence 

For real roots 

Hence

i.e.

For deuteron-proton scaltering  

 6.252. Assuming the radius of a nucleus to be equal to  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

where 

Then the number of nucleous per unit volume is  

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

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

(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

(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

 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

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

Thus the nucleus is Be^8. 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

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

(since on removing a neutron from B¹¹ we get 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 )

= - 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)]

= 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

= 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 

(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  

 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

6.260. The nuclei involved in the nuclear reaction A₁ + A₂ → → A_{3 +} A₄ have the binding energies E₁, E₂, E₃, and E₄. Find the energy of this reaction.

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

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

Now

Z₁ + Z₂ = Z₃ + Z_{4 (}conservation of change)

A₁ + 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

 6.023 x 10²³ x 200MeV = 8.21 x 10¹⁰kJ

The mass of coal with equivalent calorific value is

= 2.74 x 10⁶ kg

(b) The required mass' is

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).

 Then 

= 0.02820-0.00260 
= 0.02560 amu = 23.8 MeV 

Hence the energy released in 1 gm of He⁴ is

 23.8 x 16.02 x 10^-13 Joule = 5.75 x 10⁸ kJ

This eneigy can be derived from

=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 Li⁶ + 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 

= 0.01513 + 0.01410 - 2 x 0.00 260 amu 

= 0.02403 amu = 22.37 MeV

2.796 MeV/nucleon.

This should be compared with the value = 0.85 MeV/nucleon

6.264. 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. 

Ans.  The eneigy of reaction

is,            2 x B.E. of He⁴ - B.E. of Li⁷ 

= 8εα - 7ε_Li = 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 T_α = 4.0 MeV and the proton outgoing at an angle θ = 60° to the motion direction of the alpha-particle has a kinetic energy T_p = 2.09 MeV. 

Ans. The reaction is N¹⁴(a, p)O¹⁷

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

The momentum of outgoing proton is

Where 

and m_o is the mass of O¹⁷

The momentum of O¹⁷ is

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

Hence by definition of the Q of reaction

6.266. Making use of the tables of atomic masses, determine the energies of the following reactions: (a) Li⁷ (p, n) Be⁷; (b) Be⁹ (n,γ) Be¹⁰; (c) Li⁷  (α, n) B¹⁰); (d) O¹⁶  (d, α) N¹⁴.

Ans. (a) The reaction is  and the energy of reaction is

=-1.64 MeV

(b) The reaction is 

Mass of γ is taken zero. Then

= (0.01219 + 0.00867 - 0.01354) 

= 6.81 MeV

(c ) The reaction is  Li² (α,n) B¹⁰. The energy is

= (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

= (- 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

= (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, v_a and v_Li

then 4v_a = 7v_Li

and = 2.79 x 1.602 x 10^-6

So  2.70 x 10¹⁸ cm²/s² 

or         v_a = 9.27x10⁶ m/s

Then        v_Li = 5.3 x 10⁶ m/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 P_p = P_Be (since initial Li and final neutron are both at rest)

Then  

 Hence   = 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 

The momentum of C¹² is

Then by energy conservation

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  Li⁷ (p, α) He⁴ is

= (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

(p_p is the momentum of proton). Also

Hence 

Hence  

Substitution gives            θ = 170.53°

Also