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

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**Q.249. An alpha-particle with kinetic energy T _{α} =7 .0 MeV is scattered elastically by an initially stationary Li^{6} nucleus. Find the kinetic energy of the recoil nucleus if the angle of divergence of the two particles is **

**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 10^{11}kg/cc

**6.253. Write missing symbols, denoted by x, in the following nuclear reactions: (a) B ^{10} (x, α) Be^{8} **

(b) O^{17 } (d, n) x; (c) Na^{23} (p, x) Ne^{20}; (d) x (p, n) Ar^{37}.

**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 O^{17}. 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 A1 ^{27} 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 O ^{16} nucleus; (b) the binding energy of a neutron and an alpha-particle in a B^{11} nucleus; (c) the energy required for separation of an O^{16} nucleus into four identical particles. **

**Ans. **(a) Total binding eneigy of Ihe O^{16} 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 B^{11} nucleus

(since on removing a neutron from B^{11} we get B^{10})

= 0.01231 amu = 11.46 MeV

B.E. of ( an α-particle in B^{11})

= B.E. of B^{1} - B.E. of Li^{7} - B.E. of α

(since on removing an a from B^{11 }we get Li^{7} )

= - 0.00930 + 0.01601 + 0.00260

= 0.00931 amu = 8.67 MeV

(c) This eneigy is

[B.E. of O^{16} + 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 B ^{11} nucleus. Explain why there is the difference.**

**Ans. **B.E. o f a neutron in B^{11} - B.E. of a proton in B^{11}

= 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 Be^{10 }which together constitute B^{11}.

**6.258. Find the energy required for separation of a Ne ^{20} nucleus into two alpha-particles and a C^{12} nucleus if it is known that the binding energies per one nucleon in Ne^{20}, He^{4}, and C^{12} 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 Ne^{20} - 2 (BE. of He^{4}) - B.E. of C^{12}

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

**6.259. Calculate in atomic mass units the mass of (a) a Li ^{8} atom whose nucleus has the binding energy 41.3 MeV; (b) a C^{10} 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 C^{10} 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 C^{10} is 10.01678 amu

**6.260. The nuclei involved in the nuclear reaction A _{1 }+ A_{2} → → A_{3 + }A_{4 }have the binding energies E_{1}, E_{2}, E_{3}, and E_{4}. Find the energy of this reaction.**

**Ans. Suppose M _{1} , M_{2} , M_{3} , M_{4 }are the rest masses of the nuclei A_{1 }lf A_{2} , A_{3} and A_{4} participating in the reaction**

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

Now

Z_{1 }+ Z_{2} = Z_{3} + Z_{4 (}conservation of change)

A_{1 }+ A_{2} = A_{3} + A_{4} (conservation of heavy particles)

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

**6.261. Assuming that the splitting of a U ^{236} 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 U^{235}; (b) the mass of U^{235 }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 U^{235} is

6.023 x 10^{23 }x 200MeV = 8.21 x 10^{10}kJ

The mass of coal with equivalent calorific value is

= 2.74 x 10^{6} kg

(b) The required mass' is

**6.262. What amount of heat is liberated during the formation of one gram of He ^{4} from deuterium H^{2}? 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 He^{4} is

23.8 x 16.02 x 10^{-13} Joule = 5.75 x 10^{8} kJ

This eneigy can be derived from

=1.9 x 10^{4} 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 ^{6 }+ H^{2} → 2He^{4}. Compare the obtained magnitude with the energy per nucleon liberated in the fission of U^{235} 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 Li ^{7 }+ p → 2He^{4 }if the binding energies per nucleon in Li^{7} and He^{4} 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 He^{4} - B.E. of Li^{7 }

= 8εα - 7ε_{Li }= 8 x 7.06 - 7 x 5.60 = 17.3 MeV

**6.265. Find the energy of the reaction N ^{14 } (α, p) O^{17} 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 N^{14}(a, p)O^{17}

It is given that (in the Lab frame where N^{14} 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 O^{17}

The momentum of O^{17} 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) Li ^{7 }(p, n) Be^{7}; (b) Be^{9} (n,γ) Be^{10}; (c) Li^{7 } (α, n) B^{10}); (d) O^{16 } (d, α) N^{14}.**

**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 Li^{2 }(α,n) B^{10}. The energy is

= (0.01601 + 0.00260 - 0.00867 - .01294) amu x c^{2}

= - 2.79 MeV

(d) The reaction is O^{16} (d , α) N^{14} The energy of reaction is

= (- 0.00509 + 0.01410 - 0.00260 - 0.00307) amu x c^{2}

= 3.11 MeV

**6.267. Making use of the tables of atomic masses, find the velocity with which the products of the reaction B ^{10} (n, **α

**Ans. **The reaction is B^{10} (n, ct) Li^{1}. The energy of the reaction is

= (0.01294 + 0-00867 - 0.00260 - 0.01601) amu x c^{2}

= 2-79 MeV

Since the incident neutron is very slow and B^{10} 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 10^{18} cm^{2}/s^{2}

or v_{a} = 9.27x10^{6 }m/s

Then v_{Li} = 5.3 x 10^{6 }m/s

**6.268. Protons striking a stationary lithium target activate a reaction Li ^{7} (p, n) Be^{7}. At what value of the proton's kinetic energy can the resulting neutron be stationary?**

**Ans. **Q of this reaction (Li^{7}(p, n)Be^{7}) 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 Be ^{9} (**α

**Ans. **It is understood that Be^{9} is initially at rest. The moment of the outgoing neutron is

The momentum of C^{12 }is

Then by energy conservation

**6.270. Protons with kinetic energy T =1.0 MeV striking a lithium target induce a nuclear reaction p + Li ^{7} → 2He^{4}. 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^{7 }(p, α) He^{4 }is

= (0.01601 + 0.00783 - 0.00520) amu x c^{2}

= 0.01864 amu x c^{2} - 17.35 MeV

Since the direction of He^{4} nuclei is symmetrical, their momenta must also be equal. Let T be the K.E. of each He^{4}. Then

(p_{p} is the momentum of proton). Also

Hence

Hence

Substitution gives θ = 170.53°

Also

The document Irodov Solutions: Nuclear Reactions- 1 - Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.

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