Workout Examples No.of nucleons Binding energy

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1 Workout Examples 1. Find (i) mass defect (ii) binding energy (iii) binding energy per nucleon for a helium nucleus. Given the mass of helium nucleus= a.m.u., mass of proton= a.m.u. and mass of neutron = a.m.u. Solution: (i) Mass defect, m = [Zm p + (A Z)m n ] M = [ ] = a.m.u. (ii) Binding energy, E= mc = ( ) (iii) = J = = ev = 8.53 MeV Binding energy Binding energy per nucleon= = 7.13 MeV = 8.53 No.of nucleons Binding energy per nucleon for 9U is about 7.5 MeV, whereas it is about 8.5 MeV for nuclei of 38 half that mass. If a 9U nucleus were to split into two equal size nuclei, about how much energy would be released in the process? Solution: There are 38 nucleons involved. Each nucleon will release about =1 MeV of energy when the nucleus undergoes fission. The total energy liberated is therefor about 1 38 MeV The precise mass in the reaction 1H + 9 F He + 8 O have been determined by mass spectrometer and are m(h)= u, m(he) = u, m( 19 9 F ) = u; m( 16 8 O ) = u. Determine the Q and the nature of the reaction H + 9 F He + 8O + Q Or, Q = ( ) ( ) = u = MeV = 8.11 MeV Since Q is positive, the reaction is exoergic The Q value of the 11Na (n, α) F 0 reaction is 5.4 MeV. Determine threshold energy of the neutrons for this reaction. Given mass of Na 3 is.9898 u, Solution: The reaction is 0n Na 9 F + He + Q, where Q= 5.4 MeV, mass of incident particle i.e. of neutron( 1 0 n), m i = u, mass of target sodium, m t =.9898 u. Now threshold energy E th = Q ( m i+m t ) = +5.4 ( ) = MeV. m t A nuclear reaction is given by 7 N + 0 n 6C + 1H MeV. Find the mass of 6 C in amu. Solution: We have reaction N + 0 n 6C + 1H MeV. Or, u u m(c 14 ) u u 931 m(c 14 ) = u. 6. How much energy must a photon have if it is to split an α particle into a triton ( 3 1 H) and a proton? m( 3 1 H) 4 = u, m( He) = u, m( 1 1 H) = u [18.8 MeV] Solution: The reaction occurs is He + γ 1 H + 1 H

2 The Q value is 4 [m( He) { m( 3 1 H) + m( 1 1 H)}] 931MeV Q= [ u ( u u)] 931 = MeV The negative value of Q indicates that the reaction is endoergic. 7. A reactor is developing energy at the rate of 3000 kw. How many atoms of U 35 undergo fission per second? How many kilograms of U 35 would be used in 1000 hours of operation assuming that on an average energy of 00 MeV is released per fission? Solution: Energy released in a fission = 00 MeV= J Energy emitted by the reactor per second= 3000 kw= joule Then, number of atoms of U 35 takes part in fission reaction in one second = Number of atoms of U 35 taking part in fission reaction in 1000 hours= second = Now, according to Avogadro s hypothesis, atoms of U 35 weight 35 kg Weight of of atoms of U 35 = = kg. 8. A city require 100 Megawatts of electrical power on the average. If this is to be supplied by a nuclear reactor of efficiency 9% using U 35 as a nuclear fuel, calculate the amount of fuel required for 1 day s operation. Given that the energy released per fission of U-35 nuclide is 00 MeV. Solution: amount of energy required for 1 day operation = joule According to question, 00MeV(= J) is emitted by 1 atom of U is emitted by = atom Since only 9 % efficiency of nuclear reactor, we need atom According to, Avogadro s hypothesis atoms contain in 35 kg atoms contains in = kg The fusion reaction 1 H He + energy, is proposed to be used for the production of industrial power. Assuming the efficiency of the process to be 30%, find how many kg of deuterium will be consumed in a day for an output of kw. Given mass of 1 H = u, mass of He = u. 4 Solution: We have reaction 1 H He + energy Mass defect m =.01410u u = 0.056u. Hence energy released = But efficiency of the process is only 30%. Hence we have total energy released = 7.15 MeV. Actual output per deuterium atom is then MeV = Joule. Now output power consumed per second= Joule. Output power in a day is now Joule. Total number of deuterium required= = According to, Avogadro s hypothesis atoms contain in 35 kg atoms contain in =.94 kg.

3 10. Determine the spin parity of 4 9 Be. Solution: For odd mass number nuclei, there is either proton number or neutron number is odd. The spin parity in this case is given by P = {j, ( 1) l } where j is the total angular momentum quantum number of last unpair nucleon, l is the orbital quantum number of that nucleon. Now in our case, number of neutron is equal to 5 and hence last nucleon is a neutron which lies on 1p 3/ energy level. So for p level, l = 1, j = 3/ spin parity = [j, ( 1) l ] = [ 3, ( 1)l ] spin parity = [ 3, ] 11. Determine the spin parity of 14 7 N. Solution: Here number of proton is 7 and number of proton is 7. There is an unpair neutron and an unpair proton. Hence both these nucleon give spin parity. Now this the even mass number case of nuclei. So we use the concept of Northumb number i.e. N = j 1 + j l 1 l = 0, ±1 Case I When N=0 then, we use strong rule i.e. j = j 1 j and Spin parity = [j, ( 1) l 1+l ] Case II When N= ±1 then we use soft rule i.e. j = (j 1 + j ) or (j 1 j ) Spin parity = [j, ( 1) l 1+l ] Here seventh proton lie on the level 1p 1/, where l 1 = 1 and j 1 = 1/. This is also same for the seventh neutron. Thus j = 1/ and l = 1 Hence Northumb number, N= j 1 + j l 1 l = = 1 Here N = 1. So we use soft rule or weak rule j = (j 1 + j ) or (j 1 j ) = ( )or (1 1 ) = 1 or 0 parity = j, ( 1) l 1+l = [ 1, ( 1) 1+1 ] or [0, ( 1) 1+1 ] = ( 1, +) or (0, +)

4 1. In an absorption experiment with 1.14 MeV gamma radiation from Zn 65 it is found that 5 cm of Al reduces the beam intensity to %. Calculate the half thickness and the mass attenuation coefficient of Al for this radiation. Density of Al = 700 kg m 3. Solution: Use I = I 0 e μx, given: x = 0.5 m and I =. I Hence, we get: μ = m 1. For half thickness, x = x 1/ and I = 1. Thus x I 0 1/ = log e = μ m. Mass attenuation coefficient μ m = μ = = ρ 700 m kg. 13. It is found that 0 cm thick sheet of aluminium placed in the path of 1.1 MeV radiation beam reduces the intensity to 0%. Calculate the mass absorption coefficient of aluminum for the radiation. Density of Al = 700 kg m 3. Hint: Same as problem (1). Ans: μ m = m. kg 14. The linear attenuation coefficient for MeV γ-rays in water is about 5m 1. (i) How far must such a beam travel in water before its intensity is reduced to 1% of its original intensity? (ii) Find the relative intensity of a beam of MeV gamma rays after it has passed through 10 cm of water. Hint: (i) Use I = I 0 e μx, μ = 5 and x =? when I = 1. Ans: x = 0.91m. I (ii) x = 10 cm = 0.1 m, I =?. Solving I = I I 0 e μx, with, μ = 5 m 1, we get: I = I An electron and a positron with negligible kinetic energy annihilate each other to produce two photons. Calculate their energy, frequency and wavelength. Solution: Rest mass energy of electron E = m 0 c = ( ) = J. As an electron and a positron annihilate two produce two photons, each photon has energy equal to J. E = hν, where ν is called the frequency of the emitted photons. ν = E = ( ) = 1. h hz. Wavelength λ = c ν = = m. 16. An electron-positron pair is produced by a γ-ray of.6 MeV. How much kinetic energy is imparted to each of the charged particles. Hint: The rest mass energy of the electron-positron pair =m 0 c = ( ) = J = MeV = 1.0 MeV kinetic energy of the pair =.6 MeV 1.0 MeV = 1.4 MeV. This energy is equally shared by the electron and positron. Thus kinetic energy imparted to each particle= 1.4 = 0.6 MeV. 17. An electron and a positron each having an initial energy of 1 MeV annihilate to form two γ-ray photons. Calculate the energy of each photon. Mass of an electron and a positron is kg. Hint: Total energy of an electron= rest mass energy + 1 MeV = energy of positron = = 1.51 MeV Energy of each photon= MeV. 18. How much energy must a γ-ray photon have to produce a proton an anti-proton each having kinetic energy 10 MeV? m p = a.m.u.

5 Hint: Total energy required by a γ ray photon= total energy of the proton(or anti-proton) = [ (a. m. u) 931 MeV + 10 MeV] = MeV. 19. Photons of energy 1.0 MeV undergo Compton scattering through Calculate the energy of the scattered photon. Solution: Energy of photon, E = hν = 1.0 MeV = ν = c = E hc.thus λ = = λ h E = m Now increase in wavelength due to Compton scattering is λ = h (1 cos θ). m 0 c For θ = 180 0, we get: λ = h = = m 0 c meter. Thus the wavelength of scattered photon= ( ) = m Hence energy of scattered photon E = hν = hc = Joule = 0.04 MeV. λ Alpha particle of energy 5 MeV pass through an ionization chamber in the rate of 10 per second. Assuming all the energy is used to produce ion pairs. Calculate the current produced.(35 ev is required for producing an ion pair and e = C). Hint: Energy of alpha particles= ev. Energy required for producing one ion pair= 35eV. Number of ion pair produced= = Total number of ions produced per second, N = t = = A. Current I = Ne t 1. An ionization chamber is connected to an electromagnet of capacitance 0.5 pf and voltage sensitivity of 4 divisions per volt. A beam of alpha particles causes a deflection of 0.8 divisions. Calculate the number of ion pairs required and the energy of the alpha particles. Given that 35 ev is required for producing an ion pair and e = C. Hint: Voltage sensitivity of electrometer= 4 divisions/volt Voltage required to produce a deflection of 0.8 divisions= 0.8 = 0. volt. The total charge required Q = CV = 0.5 pf 0. = = C No. of ion pairs required= = Total energy required= 35 ( ) ev = 1.88 MeV.. It is required to operate a proportional counter with a maximum radial field of 10 7 V/m. What is the applied voltage required if the radii of the wire and tube are 0.00 cm and 1 cm respectively. Hint: Radial field E at distance r from the centre is given by V E = r log e (b/a ) For the maximum field at the wire, r = a, where a is the radius of the wire and b is the radius of the tube. E = V a log e (b/a ) V 10 7 = 10 5 log e ( ) Or, V = 14 volts. 3. A self-quenched G-M counter operates at 1000 volts and has a wire diameter of 0. mm. The radius of the cathode is cm and the tube has a guaranteed life time of 10 9 counts. What is 4

6 the maximum radial field and how long the counter last if it is used on an average for 30 hours per week at 3000 counts per minute? Consider 50 weeks to a year. Hint: Radial field E at distance r from the centre is given by V E = r log e (b/a ) For the maximum field at the wire, r = a, where a is the radius of the wire and b is the radius of the tube. E = V a log e (b/a ) 1000 Or, E = log e ( 10 / ) = volts/meter. If the life time of the tube is N years, then total number of counts recorded in its life time will equal to 10 9, i.e. N = 10 9 N= 3.7 years. 4. The wire in a GM counter collects electrons per discharge. If the count rate is 1000/minute, calculate the average current in the circuit. Hint: current I = ne = t = Amp. 5. Measurements on the track of a particle carrying a charge e in a detector give the value of 330 MeV/c for its momentum and 15 MeV for its kinetic energy. Find the rest mass of the particle. If the detector is kept in a magnetic field of 1 T and the particle moves normal to the field, what is the curvature of the track in the detector? Solution: Given momentum, p = 330 MeV, where c is the velocity of light. Kinetic energy E = c 15 MeV. We have the relation for kinetic energy of the charge particle E = [p c + (m 0 c ) ] m 0 c, where m 0 c is the rest mass energy of the particle. Say E 0 = m 0 c. Then E = [p c + E 0 ] E 0. Putting the values of E and p we get: 15 = (330) + E 0 E 0 E 0 = = MeV 15 Again B = 1 T, radius of curvature (R) =? We know Bqv = mv mv R = = p = 330MeV = R Bq Bq c Be = 1.1 m 6. In a linear accelerator, proton accelerated thrice by a potential of 40 kv leaves a tube and enters an accelerating space of length 30 cm before entering the next tube. Calculate the frequency of the r.f. voltage and the length of the tube entered by the proton. Given, e = m C kg. Solution: The kinetic energy acquired by the proton emerges out of the n th tube is given by: 1 mv n = nve Since the proton is accelerated thrice before leaving a tube say third, then 1 mv 3 = 3Ve = e, where v 3 is the velocity of proton entering the accelerating space between the tube say third and fourth. v 3 = = m s.

7 Let v 4 be the velocity of the proton leaving the accelerating space between third and fourth tube then v 4 = = m/s Mean velocity while traveling the accelerating space of length 0.3m is given by: v = v 3+v 4 = m/s. The time taken to travel 0.3 m is only the half the time period ( T ) of the r.f. voltage. time ( T ) = distance = 0.3 velocity frequency of the r.f. voltage= = Hz. Length of the next tube (fourth tube) entered by the proton(l)= velocity time = v 4 T = = m. 7. A cyclotron in which the flux density is 1.4 weber/m is employed to accelerate protons. How rapidly should the electric field between the dees be reversed? Mass of the proton= kg, charge e = C. Solution: B = 1.4 weber/m ; m = kg and e = C. Time taken by the proton to travel the semicircular path, t = πm, which is equal to the time for reversing the field. Be t = πm = = sec. Be Deuteron in a cyclotron describe a circle of radius 0.3 m just before emerging from the dees. The frequency of the applied e. m. f. is 10 Mhz. Find the flux density of the magnetic field and the velocity of the deuterons emerging out of the cyclotron. Mass of deuteron= kg, e = C. Solution: radius r = 0.3 m, frequency f = 10 Mhz = hz, mass m = kg, B =? v =? We have frequency of the applied e.m.f. is given by f = Be πmf B = = πm e = 1.30 weber/m Also mv = Bev v = Ber max = = m/s. r max m Calculate the average energy per turn and the final energy of an electron in a betatron. Solution: Suppose that magnetic flux in the betatron is given by the relation φ = φ 0 sin ωt As an increasing magnetic flux in a given direction is only obtained during the quarter cycle in which the current in the electromagnet increases from zero to maximum value. Hence time of acceleration= T = 1 π = π, where T is the time period of the changing 4 4 ω ω magnetic flux and ω is the corresponding angular frequency. Energy gained by the electron per turn, ε = Ee, where E is the induced electric field due to change in magnetic field, given by E = dφ = e d (φ d dt dt 0 sin ωt) = eφ 0 (sin ωt) dt This is the energy gained in time T = π. 4 ω

8 Total energy Hence average value of energy per turn= Time eφ 0 ω π. But φ 0 = πr B. = π/ eφ d 0 (sin 0 ωt)dt dt π ω Hence average energy per turn = eω πr B = 4eωr B π The total distance travelled in the accelerating time = c T = c π. 4 4 ω Total number of revolutions= N = (c 4 π ω ) πr = c 4ωr Let ε be the final energy acquired by the electrons, then ε = Number of revolutions made Average energy per revolution = c 4ωr 4eωr B = Berc π = eφ 0ω(sin ωt) 0 π = 30. In a certain betatron the maximum magnetic field at orbit was 0.4 weber/m, operating at 50 Hz with a suitable orbit diameter of 1.54 m. Calculate the average energy gained per revolution and the final energy of the electrons. Solution: Applied field B = 0.4 wb/m, operating frequency f = 50 Hz, radius of the path, r = 1.54 m = 0.76 m. Total number of revolutions= N = c = = = ωr 4 πf r 8π Let ε be the final energy acquired by the electrons, then ε = Berc ε = = ( J) = MeV Average energy gained per revolution, ε av = E = = Joule = N ev. We can use average energy gained directly by the formula, ε av = E = Berc c = 4Ber ω. N 4ωr 31. A betatron working on an operating frequency of 60 Hz has a stable orbit of diameter 1.6m. Find the energy gained also the final energy if the magnetic field at the orbit is 0.5 telsa. Hint: Use average energy per turn = 4eωr B and total energy = Berc. Ans: 48.6 ev, 10 MeV. 3. Complete the following nuclear reactions: (i) 17Cl+? 16S + He (ii) B+? Li + He (iii) 3Li+? 4Be + n Ans: 1 1 H, 1 0 n, 1 H 33. Using the baryon number and the strangeness number conservation laws, find which of the following reactions is allowed: (a) π + p Λ 0 + K 0 (b) π + p Λ 0 + π 0 Ans: since the reaction (a) satisfy both baryon number and strangeness number conservation laws, it is allowed. But reaction (b) is not allowed because it only satisfy baryon number conservation law but not strangeness number. 34. Using the law of conservation of lepton numbers, find which of the following reactions is possible: (a)p + ν e n + μ + (b) p + ν e n + e +

9 Ans: Since both ν e and e + have lepton number +1, the reaction (b) is possible. For the reaction (a) : p + ν e n + μ + L e : (0) + (+1) Hence reaction (a) is not possible. 35. Which of the following reactions can occur? State the conservation laws violated by the others. (a)p + p n + p + π + (b)p + p p + Λ 0 + Σ + (c) e + + e + μ + + π (d) Λ 0 π + + π (e) π + p n + π 0 Solution: For the reaction (a) p + p n + p + π + L = conserved B= conserved Y = conserved. S = conserved I 3 = conserved q = conserved Since all the quantum number are conserved, the reaction can occur. For the reaction (b) Baryon number B, hyper charge (Y), Strangeness (S) and I 3 are not conserved, the reaction cannot occur. In reaction (c), e-lepton (L e ) and μ Lepton(L μ ) are not conserved, the reaction can not occur. Similarly the reaction (d) also not allowed because the baryon number (B) and strangeness number (S) are not conserved. However the reaction (e) can occur, because all the quantum numbers are conserved as follows: π + p n + π 0 L = conserved B= conserved Y = conserved. S = conserved I 3 = conserved q = conserved 36. What particles correspond to quark composition us, ddu, sss, uus, ds and uds. Solution: (i) us corresponds to K + hadron. It has a charge q = = 1, spin( ) is zero and 3 3 strangeness S = = 1. (ii) ddu corresponds to neutron (n) : It has charge q = = 0, spin ( ) 1, strangeness S = = 0 (iii)sss represents to Ω. It has charge q = = 1, spin( ) 3, strangeness S = = 3. (iv) (v) uus represents Σ + hadron. It has charge q = = +1, Spin ( ) 3 and strangeness S = = 1. ds represents K 0 hadron. It has charge q = = 0. Spin( ) = 0 and strangeness 3 3 S = = 1.

10 (vi) uds represents Σ 0 hadron. It has charge q = = 0, spin( ) = 1 and strangeness S = = The half-value period of radium is 1590 years. In how many years will one gram of pure element (a) lose one centi-gram, and (b) be reduced to one centigram? Solution: Half life period of radium= T 1/ = 1590 years. Radioactive constant= λ = (a) Let t be time in which one gram of radium loses one centigram (0.01g). Radium left behind = = 0.99 gram. Now, decay equation is given by N = N 0 e λt log e N = log e N 0 λt Or, t = 1 log λ e ( N 0 ) = 1590 log N e ( 1 ) = 3.5 years (b) Here, N = 0.01 gram, N 0 = 1 gram = 0.01 kg; t =? t = 1 log λ e ( N 0 ) = log N e ( N 0 ) = 1590 log N e ( 1 ) = years Calculate the time required for 10% of a sample of thorium to disintegrate. Assume that the half-life of thorium to be years. Solution: We have, N = N 0 e λt. According to question, N = 0.9 N 0, t =? λ = = T 1/ N = N 0 e λt t = 1 log λ e ( N 0 ) = log N e ( N 0 ) = years. 0.9 N gram of a radioactive substance disintegrates at the rate of disintegrations per second. The atomic weight of the substance is 6. Calculate its mean life. Solution: Given Number of atoms disintegrated per second = , N = 1 gram. The mass of the substance disintegrated in one second= = kg dn kg = = λn dt sec λ = dn dt = = s 1 N Mean life T = 1 = 7.4 λ 1011 sec = 97.81years. 40. Calculate the weight in kg of one curie of Ra B(Pb 14 ) from the half life of 6.8 minutes. Solution: 1 curie = disintegrations/ sec. T 1/ = 6.8 minutes = sec. dn dn = 3.7 dt 1010 dt = λn N = = = atoms. λ atoms are contained in 14 kg of Ra B atoms are contained in kg. = kg 41. Find the activity of 1mg(10 6 kg) of radon from the half-life 3.8 day. Solution: The activity dn dt =? T 1/ = sec. λ = = s 1

11 The number of atoms in 10 6 kg of radon, N = 10 6 ( ) = The activity dn dt = λn = = events/sec= becquerel(bq).

Multiple Choice Questions

Multiple Choice Questions Nuclear Physics & Nuclear Reactions Practice Problems PSI AP Physics B 1. The atomic nucleus consists of: (A) Electrons (B) Protons (C)Protons and electrons (D) Protons and neutrons (E) Neutrons and electrons