Irodov Solutions: Elementary Particles- 1 Notes | EduRev

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

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Q.291. Calculate the kinetic energies of protons whose momenta are 0.10, 1.0, and 10 GeVic, where c is the velocity of light. 

Ans. The formula is

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Thus         T = 5.3 MeV for      Irodov Solutions: Elementary Particles- 1 Notes | EduRev

T = 0.433 GeV forIrodov Solutions: Elementary Particles- 1 Notes | EduRev

T = 9.106 GeV for   Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Here we have used mo c2 = 0.938 GeV 

 

Q.292. Find the mean path travelled by pions whose kinetic energy exceeds their rest energy η = 1.2 times. The mean lifetime of very slow pions is ζo = 25.5 ns. 

Ans. Energy of pions is (1 + η) mo c2 so

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Hence  Irodov Solutions: Elementary Particles- 1 Notes | EduRev   or  Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Here  Irodov Solutions: Elementary Particles- 1 Notes | EduRev of pion. Hence time dilation factor is 1 + η and the distance traversed by the pion in its lifetime will be

Irodov Solutions: Elementary Particles- 1 Notes | EduRevIrodov Solutions: Elementary Particles- 1 Notes | EduRev15.0 metres

on substituting the values of various quantities. (Note. The factor  Irodov Solutions: Elementary Particles- 1 Notes | EduRev  can be looked at as a time dilation effect in the laboratory frame or as length contraction factor brought to the other side in the proper frame of the pion).

 

Q.293. Negative pions with kinetic energy T = 100 MeV travel an average distance l = 11 m from their origin to decay. Find the proper lifetime of these pions. 

Ans. From the previous problem

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

where  Irodov Solutions: Elementary Particles- 1 Notes | EduRev  is the rest mass of pions.

substitution gives Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

where we have used  Irodov Solutions: Elementary Particles- 1 Notes | EduRev

 

Q.294. There is a narrow beam of negative pions with kinetic energy T equal to the rest energy of these particles. Find the ratio of fluxes at the sections of the beam separated by a distance l = = 20 m. The proper mean lifetime of these pions is ζo= 25.5 ns. 

Ans. here Irodov Solutions: Elementary Particles- 1 Notes | EduRevso the life time of the pion in the laboratory frame is

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

The law of radioactive decay implies that the flux decrease by the factor.

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

 

Q.295. A stationary positive pion disintegrated into a muon and a neutrino. Find the kinetic energy of the muon and the energy of the neutrino.

Ans. Energy-momentum conservation implies

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

But     Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Hence Irodov Solutions: Elementary Particles- 1 Notes | EduRev

So  Irodov Solutions: Elementary Particles- 1 Notes | EduRevIrodov Solutions: Elementary Particles- 1 Notes | EduRev

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Substituting  Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Also  Irodov Solutions: Elementary Particles- 1 Notes | EduRev

 

Q.296. Find the kinetic energy of a neutron emerging as a result of the decay of a stationary ∑ - hyperon (∑ - →n + π-). 

Ans. We have

Irodov Solutions: Elementary Particles- 1 Notes | EduRev       (1)

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

or Irodov Solutions: Elementary Particles- 1 Notes | EduRev

or Irodov Solutions: Elementary Particles- 1 Notes | EduRev

because (1) implies      Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Hence  Irodov Solutions: Elementary Particles- 1 Notes | EduRev

and    Irodov Solutions: Elementary Particles- 1 Notes | EduRevIrodov Solutions: Elementary Particles- 1 Notes | EduRev

Substitution gives T= 19.55 MeV 

 

Q.297. A stationary positive muon disintegrated into a positron and two neutrinos. Find the greatest possible kinetic energy of the positron. 

Ans.  The reaction is

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

 The neutrinoes are massless. The positron will carry largest momentum if both neutrionesIrodov Solutions: Elementary Particles- 1 Notes | EduRev move in the same direction in the rest frame of the nuon. Then the final product is effectively a two body system and we get from problem (295)

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Substitution gives  Irodov Solutions: Elementary Particles- 1 Notes | EduRev

 

Q.298. A stationary neutral particle disintegrated into a proton with kinetic energy T = 5.3 MeV and a negative pion. Find the mass of that particle. What is its name? 

Ans. By conservation of energy-momentum

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Then Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

This is a quadratic equation in M Irodov Solutions: Elementary Particles- 1 Notes | EduRev

or using Irodov Solutions: Elementary Particles- 1 Notes | EduRev and solving 

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Hence, Irodov Solutions: Elementary Particles- 1 Notes | EduRev

taking the positive sign. Thus

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Substitution gives

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

From the table of masses we identify the particle as a ∧ particle

 

Q.299. A negative pion with kinetic energy T = 50 MeV disintegrated during its flight into a muon and a neutrino. Find the energy of the neutrino outgoing at right angles to the pion's motion direction.

Ans. See the diagram. By conservation of eneigy

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

or

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

orIrodov Solutions: Elementary Particles- 1 Notes | EduRev

Hence the eneigy of the neutrino is

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

on writing Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Substitution gives Ev = 21.93 MeV

 

Q.300. A ∑+ hyperon with kinetic energy T∑ = 320 MeV disintegrated during its flight into a neutral particle and a positive pion outgoing with kinetic energy Tπ = 42 MeV at right angles to the hyperon's motion direction. Find the rest mass of the neutral particle (in MeV units). 

Ans. By eneigy conservation

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

or Irodov Solutions: Elementary Particles- 1 Notes | EduRev

or Irodov Solutions: Elementary Particles- 1 Notes | EduRevIrodov Solutions: Elementary Particles- 1 Notes | EduRev

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

or using the K.E. of ∑ & π

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

andIrodov Solutions: Elementary Particles- 1 Notes | EduRev

 

Q.301. A neutral pion disintegrated during its flight into two gamma quanta with equal energies. The angle of divergence of gamma quanta is Irodov Solutions: Elementary Particles- 1 Notes | EduRevFind the kinetic energy of the pion and of each gamma quantum.

Ans. Here by conservation of momentum

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Thus Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

or  Irodov Solutions: Elementary Particles- 1 Notes | EduRev

and  Irodov Solutions: Elementary Particles- 1 Notes | EduRev

substitution gives  Irodov Solutions: Elementary Particles- 1 Notes | EduRev

 Also

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

= mx c2 in this case (θ = 60°)

 

Q.302. A relativistic particle with rest mass m collides with a stationary particle of mass M and activates a reaction leading to formation of new particles: m + M →m1 +m2 . ..,where the rest masses of newly formed particles are written on the right-hand side. Making use of the invariance of the quantity E2  — p2c2 , demonstrate that the threshold kinetic energy of the particle m required for this reaction is defined by Eq. (6.7c). 

Ans. With particle masses standing for the names of the particles, the reaction is

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

On R.H.S. let the energy momenta beIrodov Solutions: Elementary Particles- 1 Notes | EduRevetc. On the left the energy momentum o f the particle m is Irodov Solutions: Elementary Particles- 1 Notes | EduRev and that of the other particle is Irodov Solutions: Elementary Particles- 1 Notes | EduRevwhere, of course, the usual relations

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

hold. From the conservation of energy momentum we see that

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Left hand side is

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

We evaluate the R.H.S. in the frame where  Irodov Solutions: Elementary Particles- 1 Notes | EduRev (CM frame of the decay product).

Then          Irodov Solutions: Elementary Particles- 1 Notes | EduRev

because all energies are +ve. Therefore we have the result

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

or  since E = mc2 + T, we see that  T ≤ Tth where

 Irodov Solutions: Elementary Particles- 1 Notes | EduRev

 

Q.303. A positron with kinetic energy T = 750 keV strikes a stationary free electron. As a result of annihilation, two gamma quanta with equal energies appear. Find the angle of divergence between them.

Ans. By momentum conservation

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

or  Irodov Solutions: Elementary Particles- 1 Notes | EduRev Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Substitution gives

θ = 98.8°

 

Q.304. Find the threshold energy of gamma quantum required to form (a) an electron-positron pair in the field of a stationary electron; (b) a pair of pions of opposite signs in the field of a stationary proton.

Ans. The formula of problem 3.02 gives

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

when the projectile is a photon 

(a) For Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

(b) ForIrodov Solutions: Elementary Particles- 1 Notes | EduRev

Irodov Solutions: Elementary Particles- 1 Notes | EduRevIrodov Solutions: Elementary Particles- 1 Notes | EduRevIrodov Solutions: Elementary Particles- 1 Notes | EduRev

 

Q.305. Protons with kinetic energy T strike a stationary hydrogen target. Find the threshold values of T for the following reactions: (a) Irodov Solutions: Elementary Particles- 1 Notes | EduRev (b) Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Ans. (a) For Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

(b) For Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

 

Q.306. A hydrogen target is bombarded by pions. Calculate the threshold values of kinetic energies of these pions making possible the following reactions:   (a)Irodov Solutions: Elementary Particles- 1 Notes | EduRev  (b) Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Ans. (a) Here

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Substitution gives Tth = 0.904 GeV 

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Substitution gives Tth = 0.77 GeV. 

 

Q.307. Find the strangeness S and the hypercharge Y of a neutral elementary particle whose isotopic spin projection is Tz = +1/2 and baryon charge B = +1. What particle is this? 

 Ans. From the Gell-Mann Nishijima formula

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

we get 

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Also Irodov Solutions: Elementary Particles- 1 Notes | EduRev. Thus the particle is =° 0.

 

Q.308. Which of the following processes are forbidden by the law of conservation of lepton charge: 

Irodov Solutions: Elementary Particles- 1 Notes | EduRev   Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Ans. (1) The process Irodov Solutions: Elementary Particles- 1 Notes | EduRev  cannot occur as there are 2 more leptons Irodov Solutions: Elementary Particles- 1 Notes | EduRev on the right comopared to zero on the left

(2 ) The process Irodov Solutions: Elementary Particles- 1 Notes | EduRev is forbidden because this corresponds to a change of lepton number by, (0 on the left - 1 on the right)

(3) The process Irodov Solutions: Elementary Particles- 1 Notes | EduRev is forbidden because Irodov Solutions: Elementary Particles- 1 Notes | EduRev being both leptons ΔL = 2hre.

(4) , (5), (6) are allowed (except that one must distinguish between muon neutrinoes and electron neutrinoes). The correct names would be

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

 

Q.309. Which of the following processes are forbidden by the law of conservation of strangeness: 

Irodov Solutions: Elementary Particles- 1 Notes | EduRev      Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Ans. (1)Irodov Solutions: Elementary Particles- 1 Notes | EduRev

SO Irodov Solutions: Elementary Particles- 1 Notes | EduRev

(2)   Irodov Solutions: Elementary Particles- 1 Notes | EduRev

SOIrodov Solutions: Elementary Particles- 1 Notes | EduRev

(3)   Irodov Solutions: Elementary Particles- 1 Notes | EduRev

SO Irodov Solutions: Elementary Particles- 1 Notes | EduRev

(4)  Irodov Solutions: Elementary Particles- 1 Notes | EduRev 

SOIrodov Solutions: Elementary Particles- 1 Notes | EduRev

(5)  Irodov Solutions: Elementary Particles- 1 Notes | EduRev

SOIrodov Solutions: Elementary Particles- 1 Notes | EduRev

(6)   Irodov Solutions: Elementary Particles- 1 Notes | EduRev

SOIrodov Solutions: Elementary Particles- 1 Notes | EduRev

 

Q.310. Indicate the reasons why the following processes are forbidden: 

Irodov Solutions: Elementary Particles- 1 Notes | EduRev     Irodov Solutions: Elementary Particles- 1 Notes | EduRev

Ans. (1) Irodov Solutions: Elementary Particles- 1 Notes | EduRev

is forbidden by energy conservation. The mass difference

Irodov Solutions: Elementary Particles- 1 Notes | EduRev

(The process 1 → 2 + 3 will be allowed only if m1 > m+ m3.)

(2) Irodov Solutions: Elementary Particles- 1 Notes | EduRev

is disallowed by conservation of baryon number.

(3) Irodov Solutions: Elementary Particles- 1 Notes | EduRev

is forbidden by conservation of charge

(4)Irodov Solutions: Elementary Particles- 1 Notes | EduRev

is forbidden by strangeness conservation.

(5)Irodov Solutions: Elementary Particles- 1 Notes | EduRev

is forbidden by conservation of muon number (or lepton number).

(6)Irodov Solutions: Elementary Particles- 1 Notes | EduRev

 is forbidden by the separate conservation of muon number as well as lepton number.

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