Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

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

JEE : Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

The document Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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 Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev = 60°. 

Ans. Initial momentum of the a particle is  Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRevis a unit vector in the incident direction). Final momenta are respectively Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev Conservation of momentum reads

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Squaring Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev(1)

where Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev is the angle between  Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Also by energy conservation Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

(m & M are respectively the masses of a particle and  Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev) So

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev(2)

Substracting (2) from (1) we see that

Thus if  Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

This being understood, we write

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Hence the recoil energy of the Li, nucleus is

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

As we pointed out above  Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

or since pd = 0 in a head on collisions

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Going back to energy conservation

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

So   Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Then by eneigy conservation

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

The energy lost by neutron in then

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

or fraction of energy lost is Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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 

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

From energy conservation 

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

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

So Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Hence Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

For real roots Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

HenceNuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

i.e.Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

For deuteron-proton scaltering  Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

 

 6.252. Assuming the radius of a nucleus to be equal to Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

where  Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Then the number of nucleous per unit volume is Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev  Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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 )

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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)]

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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 

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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  Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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 

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

NowNuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev 6.023 x 1023 x 200MeV = 8.21 x 1010kJ

The mass of coal with equivalent calorific value is

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev= 2.74 x 106 kg

(b) The required mass' is

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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).

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

 Then 

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

= 0.02820-0.00260 
= 0.02560 amu = 23.8 MeV 

Hence the energy released in 1 gm of He4 is

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev 23.8 x 16.02 x 10-13 Joule = 5.75 x 108 kJ

This eneigy can be derived from

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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 Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

= 0.01513 + 0.01410 - 2 x 0.00 260 amu 

= 0.02403 amu = 22.37 MeV

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev2.796 MeV/nucleon.

This should be compared with the value Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Where Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

and mis the mass of O17

The momentum of O17 is

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Hence by definition of the Q of reaction

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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 Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev and the energy of reaction is

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

=-1.64 MeV

(b) The reaction is Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Mass of γ is taken zero. Then

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

= (0.01219 + 0.00867 - 0.01354) Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

= 6.81 MeV

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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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 Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev= 2.79 x 1.602 x 10-6

So Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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)

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Then  Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

 Hence   Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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 

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRevThe momentum of C12 is

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Then by energy conservation

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE 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

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

(pp is the momentum of proton). Also

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Hence Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Hence  Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

Substitution gives            θ = 170.53°

Also Nuclear Reactions (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev

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