Q.1.(a) Two stable isotopes of 3Li6 and 3Li7 have respective abundances of 7.5 and 92.5 . These isotopes have masses 6.01512 u and 7.01600 u respectively. Find the atomic weight of lithium.(b) Boron has two stable isotopes 5B10. Their respective masses are 10.01294 u and 11.00931 u, and the atomic weight of boron is 10.811 u. find the abundances of 5B10 and 5B11.
Q.2. The three stable isotopes of neon 10Ne20, 10Ne21 and 10Ne22 have respective abundances of 90.51 0.27 and 9.22 . The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u respectively. Obtain the average atomic mass of neon.
Sol. The masses of there isotopes are 19.99 u, 20.99 u, 21.99 u
Q.3.Obtain the binding energy (in Me V) of a nitrogen nucleus (7N14). Given, m 7N14=14.00307 u
S0l. 7Ne14 nucleus contains 7 protons and 7 neutrons.
Q.4.Obtain the binding energy of the nuclei 26Fe56 and 83Bi209 in units of MeV from the following data : (26Fe56)=208.980388 a. m. u.Which nucleus has greater binding energy per nucleon ? Take 1 a. m. u. = 931.5 MeV
Q.5.A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity, assume that the coin is entirely made of 29Cu63 atoms (of mass 62.92960 u).
Q.6.Write nuclear reaction equation for(i) a decay of 88Ra226 (ii) a decay of 94Pu242 (iii) β – decay of 15P32
β – decay of 83Bi210 (v) β+ – decay of 6C11 (vi) β+ – decay of 43C97
(vii) Electron capture of 54Xe120 .
Q.7.A radioactive isotope has half life of T years. How long will it take the activity to reduce to (a) 3.125% (B) 1% of its original activity ?
Q.8.The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbo. This activity arises from the small proportion of radioactive 14C6 present with the ordinary 6C12 isotope. When the organism is dead, its interaction with the atmosphere which maintains the above equilibrium activity, ceases and its activity begins to drop. From the know half life (= 5730 years) of 6C14, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of 6C14 dating used in archaeology. Suppose a specimen from Mohenjodaro given an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus Valley Civilization.
Sol. Here, normal activity, R0 = 15 decays/min
Present activity R = 9 decays/min, T = 5730 Yrs, Age, t = ?
As activity is proportional to the number of radioactive atoms, therefore,
Present activity R = 9 decays/min, T = 5730 Yrs, Age, t = ?
As activity is proportional to the number of radioactive atoms, therefore,
Q.9.Obtain the amount of 27Co60 necessary to provide a radioactive source of 8.0 m Ci strength. Half- life of 27Co60 is 5.3 years.Sol. Here, mass of 27Co60 =?
Q.10. The half life of 38Sr90 is 28 years. What is the disintegration rate of 15 mg of this isotope ?
Q.11. Obtain approximately the ratio of the nuclear radii of the gold isotope 79Au197 and silver isotope 47Ag107.
Q.12.. Find the Q value and the kinetic energy of emitted a particle in the a decay of(a) 88Ra226 (b) 86Rn220. Given m (88Ra26) = 226.02540 u ; m = (86Rn222 = 222.01750 u
(b) m (86Rn220) = 220.01137 u ; m (86Rn216) = 216.00189 u and
Q.13.The radionuclide 6C11 decays according to 6C11 5B11 + e+ v : half-life = 20.3 min.The maximum energy of the emitted positron is 0.960 MeV. Given the mass values
6C11)=11.011434 u ; m (6B11) = 11.009305 u
Calculate Q and compare it with maximum energy of positron emitted.
Sol.Mass defect in the given reaction is m = m(6C11) – [m (5B11) +me]This is in terms of nuclear masses. If we express the Q value in terms of atomic masses, we have to subtract 6 me from atomic mass of carbon and 5 me from that of boron to get the corresponding nuclear masses. Therefore, we have
Q.14.The nucleus 10Ne23 decays by β – decay equation and determine the maximum kinetic energy of the electrons emitted from the following data :
m (10Ne23) = 22.994466 amu, m (10Ne23) = 22.989770 amu.
m (10Ne23) = 22.994466 amu, m (10Ne23) = 22.989770 amu.
Q.15. The Q value of a nuclear reaction
A + B = C + d is defined by Q = [ mA + mb – mc – md] where the masses refer to the respective nuclei. Determine from the given data the Q value the Q value of the state whether the reactions are exothermic or endothermic.
A + B = C + d is defined by Q = [ mA + mb – mc – md] where the masses refer to the respective nuclei. Determine from the given data the Q value the Q value of the state whether the reactions are exothermic or endothermic.
(i) 1H1 +1H3 1H2 +1H1
(ii) 6C12 + 6C12 10Ne20 + 2He4
Atomic masses are given to be m 1H2)= 2.014102 u ; m(1H3) = 3.016049 u ; m (6C12) = 12.00000 u ; m (10Ne20) = 19.992439 u.
Q.16. Suppose, we think of fission of a 26F e56nucleus into two equal fragments 13Al28. Is the fission energetically possible ? Argue by working out Q of the process. Given m 26F e56) = 55.93494 u, m (13Al28) = 27.98191 u.
Q.17. The fission properaties of 84Pu239 are very similar to those of 92U235. The average energy released per fission is 180 MeV. How much energy in MeV is released if all the atoms in 1 Kg of pure 94Pu239 undergo fission.
Q.18. A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much 92U235 did it contain initially ? Assume that the reactor operates 80 of the time and that all the energy generated arises from the fission of 92U235 and that this nuclide is consumed by the fission process.
Q.19. How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium? The fusion reaction can be taken as 1H2 + 1H2 2He3 + n + 3.27 MeV
Q.20. Calculate the height of potential barrier for a head-on collision of two deuterons. The effective radius of deuteron can be taken to be 2 fm. Note that height of potential barrier is given by the Coulomb repulsion between two deuterons when they just touch each other.
Sol.For head on collision, distance between centres of two deuterons = r = 2 × RadiusCharge of each deuteron e = 1.6 × 10-19 C
Q.21.From the relation R = R0 A1/3, where R0 is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e., independent of A).
Q.22. For the (positron) emission from a nucleus, there is another competing process known as electron capture. Electron from an inner orbit (say K shell ) is captured by the nucleus and a neutrino is emitted. Show that if emission is energetically allowed, electron capture is necessarily allowed but not vice-versa.
Sol.. The emission from a nucleus ZXA may be represented asZXA = Z-1yA + 1e0 + v + Q1
The other competing process of electron capture may be represented as …(i)
-1eo + ZXA = Z-1yA + v + Q2
The energy released Q1 in (i) given by ……(ii)
Q1 = [mN (ZXA) – mN (Z-1yA) – me ] C2 = [mn Z-1yA ) + Z-1yA ) – ( Z – 1 )
Q1 = [m (Z-1XA) – m (Z-1yA) – 2 me]c2 …..(iii)
Note that here denotes mass of nucleus and m denotes the mass of atom. Similarly, from (ii)
Q2 = [Mn (ZXA) + me – mN (Z-1yA)]c2
=[mN(ZXA) + Zme + m2 – mN (Z-1yA) – (Z – 1) me – me] c2
Q2 = [ m (ZXA)- m (ZXA)]
If Q1 > 0, then Q2 > 0
i.e., if positron emission is energetically allowed, electron capture is necessarily allowed.
But Q2 > 0 does not necessarily mean Q1 > 0. Hence the reverse is not true.
Q.23. In a periodic table, the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are 12Mg24 (23.98504 u), 12Mg25 (24.98584 u) and 12Mg26 (25.98259 u). The natural abundance of 12Mg24 is 78.99 by mass. Calculate the abundances of the other two isotopes.
Sol. Let the abundance of 12Mg24 by mass be . Therefore, abundance of 12Mg26 by mass
Q.24.. The neutron separation energy is defined to be the energy required to remove a neutron from a nucleus. Obtain the neutron separation energy of the nuclei 20Ca41 and 13Al27 from the followingData : m (20Ca40) = 39.962591 u and m (20Ca41) = 40.962278 u
m ( 13Al26) = 25.986895 u and m (13Al27) = 26.981541 u
∴ Neutron separation energy = 0.0138454 × 931 MeV = 12.89 MeV
Q.25. . A source contains two phosphorous radionuclides 15P32 (T1/2 = 14.3 days) and 15P33 (T1/2 = 25.3 days). Initially, 10 of the decays come from 15P33. How long one must wait until 90 do so ?
Q.26.Under certain circumstances, a nucleus can decay by emitting a particle more massive than an particle. Consider the following decay processes :88Ra223 82Pb209 + 6C14 , 88Ra223 88Rn219 + 2He4
(a) Calculate the Q-values for these decays and determine that both are energetically allowed.
AS Q values are positive in both the cases, therefore both the decays are energetically possible.
Q.27.. Consider the fission of 92U238 by fast neutrons. In one fission event, no neutrons are emitted and the final stable end products, after the beta decay of the primary fragments are 58Ca140 and 44Ru99. Calculate Q for this fission process. The relevant atomic and particle masses are :
m (92U238) = 238.05079 u ; m (58Ca140) = 139.90543 u ; m (34Ru99) = 98.90594 u
m (92U238) = 238.05079 u ; m (58Ca140) = 139.90543 u ; m (34Ru99) = 98.90594 u
Q.28. (a) Consider the so called D-T reaction (deuterium-tritium fusion) 1H2+1H3 2He4 + n Calculate the energy released in MeV in this reaction from the datam (1H2) = 2.014102 u, m (1H3) = 3.16049 u
(b) Consider the radius of both deuterium and tritium to be approximately 2-0 fm. What is the kinetic energy needed to overcome the Coulomb repulsion between the two nuclei ? To what temperature must the gases be heated to initiate the reaction ?
(b) Repulsive potential energy of two nuclei when they almost touch each other is
In actual practice, the temperature required for trigerring the reaction is some what less.
Q.29.Obtain the maximum kinetic energy of particles and the radiation frequencies of Y-decays in the decay scheme sow in You are given thatM (79Au198) = 197.968233 u,
M (80Hg198) = 197.966760 u.
Q.30. Calculate and compare the energy released by (a) fusion of 1.0 kg of hydrogen deep within the sun, and (b) the fission of 1.0 kg of U235 in a fission reactor.
Q.31. Suppose India has a target of producing by 2020 A.D., MW of electron power, ten percent of which is to be obtained from nuclear power plants. Suppose we are given that on an average, the efficiency of utilization (i.e., conversion to electrical energy) of thermal energy produced in a reactor is . How much amount of fissionable uranium will our country need per year ? Take the heat energy per fission of U235 to be about 200 MeV.
