Chapter 13 Nuclei Solutions
Question - 11 : - Obtain approximately the ratio of the nuclear radii ofthe gold isotope 
and the silver isotope 
Answer - 11 : - Nuclear radius of the gold isotope 
= RAu
Nuclear radiusof the silver isotope 
= RAg
Mass number of gold, AAu =197
Mass number of silver, AAg =107
The ratio of the radii ofthe two nuclei is related with their mass numbers as:
Hence, the ratio of the nuclearradii of the gold and silver isotopes is about 1.23.
Question - 12 : - Find the Q-value and the kinetic energy of theemitted α-particle in the α-decay of (a) 
and (b) 
.Given 
= 226.02540 u, 
= 222.01750 u, 
= 220.01137 u, 
= 216.00189 u.
Answer - 12 : - (a) Alphaparticle decay of 
emits a helium nucleus. As a result, its mass numberreduces to (226 − 4) 222 and its atomic number reduces to (88 − 2) 86. This isshown in the following nuclear reaction.
Q-value of
emitted α-particle =(Sum of initial mass − Sum of final mass) c2
Where,
c =Speed of light
It is given that:
Q-value =[226.02540 − (222.01750 + 4.002603)] u c2
= 0.005297 u c2
But 1 u = 931.5 MeV/c2
∴Q =0.005297 × 931.5 ≈ 4.94 MeV
Kineticenergy of the α-particle 
(b) Alphaparticle decay of 
is shown by the followingnuclear reaction.
It is given that:
Massof = 220.01137 u
Massof 
= 216.00189 u
≈ 641 MeV
Kineticenergy of the α-particle 
= 6.29 MeV
Question - 13 : - The radionuclide 11Cdecays according to
The maximum energy of the emittedpositron is 0.960 MeV.
Given the mass values:
calculate Q andcompare it with the maximum energy of the positron emitted
Answer - 13 : -
The given nuclear reactionis:
Atomic mass of 
= 11.011434 u
Atomicmass of 
Maximum energy possessed by theemitted positron = 0.960 MeV
Thechange in the Q-value (ΔQ) of the nuclear masses ofthe 
nucleus is given as:Where,
me =Mass of an electron or positron = 0.000548 u
c = Speedof light
m’ =Respective nuclear masses
If atomic masses are used insteadof nuclear masses, then we have to add 6 me in thecase of
and 5 me inthe case of 
.
Hence, equation (1) reduces to:
∴ΔQ =[11.011434 − 11.009305 − 2 × 0.000548] c2
= (0.001033 c2)u
But 1 u = 931.5 Mev/c2
∴ΔQ =0.001033 × 931.5 ≈ 0.962 MeV
The value of Q isalmost comparable to the maximum energy of the emitted positron.
Question - 14 : - The nucleus 
decays by 
emission. Write down the
decay equation anddetermine the maximum kinetic energy of the electrons emitted. Given that:
= 22.994466 u
= 22.989770 u.
Answer - 14 : - In emission, the number of protonsincreases by 1, and one electron and an antineutrino are emitted from theparent nucleus.
emission of the nucleus is given as: It is given that:
Atomicmass of = 22.994466 uMass of an electron, me =0.000548 u
Q-value ofthe given reaction is given as:
There are 10 electrons in and 11 electrons in
. Hence, the mass of the electron is cancelled inthe Q-value equation.
The daughter nucleus is too heavy as compared to 
and 
. Hence, it carries negligibleenergy. The kinetic energy of the antineutrino is nearly zero. Hence, themaximum kinetic energy of the emitted electrons is almost equal to the Q-value,i.e., 4.374 MeV.
Question - 15 : - The Q value of anuclear reaction A + b → C + d isdefined by
Q = [ mA+ mb− mC− md]c2 wherethe masses refer to the respective nuclei. Determine from the given datathe Q-value of the following reactions and state whether thereactions are exothermic or endothermic.
(i) 
(ii) 
Atomic masses are given to be
Answer - 15 : -
(i) Thegiven nuclear reaction is:
It is given that:
Atomicmass 
Atomic mass 
Atomic mass 
According to the question,the Q-value of the reaction can be written as:
The negativeQ-value of thereaction shows that the reaction is endothermic.
(ii) Thegiven nuclear reaction is:
It is given that:
Atomicmass of 
Atomic mass of 
Atomic mass of 
The Q-value of thisreaction is given as:
The positive Q-valueof the reaction shows that the reaction is exothermic.
Question - 16 : - Suppose, we think of fission of a 
nucleus into two equal fragments,
. Is the fission energetically possible? Argue byworking out Q of the process. Given 
and
Answer - 16 : - The fission of can be given as:
It is given that:
Atomicmass of 
= 55.93494 u
The Q-value of thisnuclear reaction is given as:
The Q-value of thefission is negative. Therefore, the fission is not possible energetically. Foran energetically-possible fission reaction, the Q-value must bepositive.
are very similar to those of
.
Theaverage energy released per fission is 180 MeV. How much energy, in MeV, isreleased if all the atoms in 1 kg of pure undergo fission?
Answer - 17 : - Average energy released per fission of, 
Amount of pure
, m = 1 kg =1000 g NA= Avogadro number =6.023 × 1023
Massnumber of= 239 g1mole of contains NA atoms.∴m gof contains
∴Totalenergy released during the fission of 1 kg ofis calculated as: Hence, 
is released if all the atoms in 1 kg ofpure undergo fission.
Question - 18 : - A 1000 MW fission reactor consumes half of its fuel in5.00 y. How much did it contain initially? Assume that the reactoroperates 80% of the time, that all the energy generated arises from the fissionof and that this nuclide is consumedonly by the fission process.
Answer - 18 : - Half life of the fuel of the fissionreactor, 
years
= 5 × 365 × 24 × 60 × 60 s
Weknow that in the fission of 1 g of nucleus, the energy released isequal to 200 MeV.1mole, i.e., 235 g of contains 6.023 × 1023 atoms.∴1 g contains
Thetotal energy generated per gram ofis calculated as:The reactor operates only80% of the time.
Hence,the amount of consumed in 5 years by the 1000MW fission reactor is calculated as:∴Initialamount of = 2 × 1538 = 3076 kg
Question - 19 : - How long can an electric lamp of100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reactionas
Answer - 19 : -
The given fusion reaction is:
Amount of deuterium, m =2 kg
1 mole, i.e., 2 g of deuteriumcontains 6.023 × 1023 atoms.
∴2.0 kg of deuteriumcontains 
It can be inferred from the givenreaction that when two atoms of deuterium fuse, 3.27 MeV energy is released.
∴Totalenergy per nucleus released in the fusion reaction:
Power of the electric lamp, P =100 W = 100 J/s
Hence, the energy consumed by thelamp per second = 100 J
The total time for which theelectric lamp will glow is calculated as:
Question - 20 : - Calculate the height of thepotential barrier for a head on collision of two deuterons. (Hint: The heightof the potential barrier is given by the Coulomb repulsion between the twodeuterons when they just touch each other. Assume that they can be taken ashard spheres of radius 2.0 fm.)
Answer - 20 : -
When two deuterons collidehead-on, the distance between their centres, d is given as:
Radius of 1st deuteron+ Radius of 2nd deuteron
Radius of a deuteron nucleus = 2fm = 2 × 10−15 m
∴d =2 × 10−15 + 2 × 10−15 =4 × 10−15 m
Charge on a deuteron nucleus =Charge on an electron = e = 1.6 × 10−19 C
Potentialenergy of the two-deuteron system:Where,
= Permittivity of freespace
Hence, the height of thepotential barrier of the two-deuteron system is 360 keV.