_{3}Li

^{6}and

_{3}Li

^{7}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

_{5}B

^{10}. 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

_{5}B

^{10}and

_{5}B

^{11}.

_{10}Ne

^{20},

_{10}Ne

^{21}and

_{10}Ne

^{22}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.

_{7}N

^{14}). Given, m

_{7}N

^{14}=14.00307 u

_{7}Ne

^{14}nucleus contains 7 protons and 7 neutrons.

_{26}Fe

^{56}and

_{83}Bi

^{209}in units of MeV from the following data : (

_{26}Fe

^{56})=208.980388 a. m. u.Which nucleus has greater binding energy per nucleon ? Take 1 a. m. u. = 931.5 MeV

_{29}Cu

^{63}atoms (of mass 62.92960 u).

_{88}Ra

^{226}(ii) a decay of

_{94}Pu

^{242}(iii) β – decay of

_{15}P

^{32}

β – decay of _{83}Bi^{210} (v) β^{+} – decay of _{6}C^{11} (vi) β^{+} – decay of _{43}C^{97}

(vii) Electron capture of _{54}Xe^{120} .

_{14}C

^{6}present with the ordinary

_{6}C

^{12}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

_{6}C

^{14}, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of

_{6}C

^{14}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.

Present activity R = 9 decays/min, T = 5730 Yrs, Age, t = ?

As activity is proportional to the number of radioactive atoms, therefore,

_{27}Co

^{60}necessary to provide a radioactive source of 8.0 m Ci strength. Half- life of

_{27}Co

^{60}is 5.3 years.Sol. Here, mass of

_{27}Co

^{60}=?

_{38}Sr

^{90}is 28 years. What is the disintegration rate of 15 mg of this isotope ?

_{88}Ra

^{226}(b)

_{86}Rn

^{220}. Given m (

_{88}Ra

^{26}) = 226.02540 u ; m = (

_{86}Rn

^{222}= 222.01750 u

(b) m (_{86}Rn^{220}) = 220.01137 u ; m (_{86}Rn^{216}) = 216.00189 u and

_{6}C

^{11}decays according to

_{6}C

^{11}

_{5}B

^{11}+ e

^{+}v : half-life = 20.3 min.The maximum energy of the emitted positron is 0.960 MeV. Given the mass values

_{6}C^{11})=11.011434 u ; m (_{6}B^{11}) = 11.009305 u

Calculate Q and compare it with maximum energy of positron emitted.

_{6}C

^{11}) – [m (

_{5}B

^{11}) +m

_{e}]This is in terms of nuclear masses. If we express the Q value in terms of atomic masses, we have to subtract 6 m

_{e}from atomic mass of carbon and 5 m

_{e}from that of boron to get the corresponding nuclear masses. Therefore, we have

_{10}Ne

^{23}decays by β – decay equation and determine the maximum kinetic energy of the electrons emitted from the following data :

m (

_{10}Ne

^{23}) = 22.994466 amu, m (

_{10}Ne

^{23}) = 22.989770 amu.

A + B = C + d is defined by Q = [ m

_{A}+ m

_{b}– m

_{c}– m

_{d}] 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) _{1}H^{1} +_{1}H^{3} _{1}H^{2} +_{1}H^{1}

(ii) _{6}C^{12} + _{6}C^{12} _{10}Ne^{20} + _{2}He^{4}

Atomic masses are given to be m _{1}H^{2})= 2.014102 u ; m(_{1}H^{3}) = 3.016049 u ; m (_{6}C^{12}) = 12.00000 u ; m (_{10}Ne^{20}) = 19.992439 u.

_{26}F e

^{56}nucleus into two equal fragments

_{13}Al

^{28}. Is the fission energetically possible ? Argue by working out Q of the process. Given m

_{26}F e

^{56}) = 55.93494 u, m (

_{13}Al

^{28}) = 27.98191 u.

_{84}Pu

^{239}are very similar to those of

_{92}U

^{235}. 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

_{94}Pu

^{239}undergo fission.

_{92}U

^{235}did it contain initially ? Assume that the reactor operates 80 of the time and that all the energy generated arises from the fission of

_{92}U

^{235}and that this nuclide is consumed by the fission process.

_{1}H

^{2}+

_{1}H

^{2}

_{2}He

^{3}+ n + 3.27 MeV

^{-19}C

_{0}A

^{1/3}, where R

_{0}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).

_{Z}X

^{A}may be represented as

_{Z}X

^{A}=

_{Z-1}y

^{A}+

_{1}e

^{0}+ v + Q

_{1}

The other competing process of electron capture may be represented as …(i)

_{-1}e^{o} + _{Z}X^{A }= _{Z-1}y^{A} + v + Q_{2}

The energy released Q_{1} in (i) given by ……(ii)

Q_{1} = [m_{N} (_{Z}X^{A}) – m_{N} (_{Z-1}y^{A}) – m_{e} ] C^{2} = [m_{n} _{Z-1}y^{A }) + _{Z-1}y^{A} ) – ( Z – 1 )

Q_{1 } = [m (_{Z-1}X^{A}) – m (_{Z-1}y^{A}) – 2 m_{e}]c^{2} …..(iii)

Note that here denotes mass of nucleus and m denotes the mass of atom. Similarly, from (ii)

Q_{2} = [M_{n} (_{Z}X^{A}) + m_{e }– m_{N} (_{Z-1}y^{A})]c^{2}

=[m_{N}(_{Z}X^{A}) + Zm_{e} + m_{2} – m_{N} (_{Z-1}y^{A}) – (Z – 1) m_{e} – m_{e}] c^{2}

Q_{2} = [ m (_{Z}X^{A})- m (_{Z}X^{A})]

If Q_{1} > 0, then Q_{2} > 0

i.e., if positron emission is energetically allowed, electron capture is necessarily allowed.

But Q_{2} > 0 does not necessarily mean Q_{1} > 0. Hence the reverse is not true.

_{12}Mg

^{24}(23.98504 u),

_{12}Mg

^{25 }(24.98584 u) and

_{12}Mg

^{26}(25.98259 u). The natural abundance of

_{12}Mg

^{24}is 78.99 by mass. Calculate the abundances of the other two isotopes.

_{12}Mg

^{24}by mass be . Therefore, abundance of

_{12}Mg

^{26}by mass

_{20}Ca

^{41}and

_{13}Al

^{27}from the followingData : m (

_{20}Ca

^{40}) = 39.962591 u and m (

_{20}Ca

^{41}) = 40.962278 u

m ( _{13}Al^{26}) = 25.986895 u and m (_{13}Al^{27}) = 26.981541 u

∴ Neutron separation energy = 0.0138454 × 931 MeV = 12.89 MeV

_{15}P

^{32}(T

_{1/2}= 14.3 days) and

_{15}P

^{33}(T

_{1/2}= 25.3 days). Initially, 10 of the decays come from

_{15}P

^{33}. How long one must wait until 90 do so ?

_{88}Ra

^{223}

_{82}Pb

^{209}+

_{6}C

^{14},

_{88}Ra

^{223}

_{88}Rn

^{219}+

_{2}He

^{4}

(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.

_{92}U

^{238 }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

_{58}Ca

^{140}and

_{44}Ru

^{99}. Calculate Q for this fission process. The relevant atomic and particle masses are :

m (

_{92}U

^{238}) = 238.05079 u ; m (

_{58}Ca

^{140}) = 139.90543 u ; m (

_{34}Ru

^{99}) = 98.90594 u

_{1}H

^{2}+

_{1}H

^{3}

_{2}He

^{4}+ n Calculate the energy released in MeV in this reaction from the datam (

_{1}H

^{2}) = 2.014102 u, m (

_{1}H

^{3}) = 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.

_{79}Au

^{198}) = 197.968233 u,

M (_{80}Hg^{198}) = 197.966760 u.

^{235}in a fission reactor.

^{235}to be about 200 MeV.