Q: Consider the nuclear fission, Ne^{20} \rightarrow 2He^4 + C^{12} Given that the binding energy/nucleon of Ne^{20} , He^4 and C^{12} are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement:

Energy absorbed; \Delta Q = \text{Binding energy of products - Binding energy of reactants} \text{Binding energy of products} = 2 \times (\text{B.E of } He^4 ) + (\text{B.E of } C^{12} ) = 2 (4 \times 7.07 ) + (12 \times 7.86) =150.88 \ MeV \text{Binding energy of Reactants} = 20 \times 8.03 = 160.6 \ MeV \Delta Q = \text{Binding energy of products - Binding energy of reactants} = 150.88 - 160.6 = -9.72 \ MeV \text{ Hence, 9.72 MeV energy will be released}

Question 1

The time taken by a particle moving with velocity  c = 10 ^{6} ms^{-1}    in crossing a nucleus will approximately be

Accepted Answer

Radius of a nucleus is given by the relation R={ R }_{ 0 }\left( { A }^{ \frac { 1 }{ 3 }  } ight) where  { R }_{ 0 }=1.2	imes { 10 }^{ -15 }m Hence, radius of the nucleus will be in order  r ={ 10 }^{ -15 }m  and so will be its diameter.Given, speed is  c=10^{6} m/s  Therefore, time taken to cross the nucleus is  t=\dfrac{r}{c} = \dfrac { { 10 }^{ -15 } }{ 10^{6} } = { 10 }^{ -21 }   second approximately.Answer is 'D'.

Question 2

The graph of  ln(R/R_{O}  ) versus  ln A  of nucleus is :( R_{0}  is a constant  1.2 	imes 10^{-12}  , R is radius and A is the atomic number)

Accepted Answer

Size of a nucleus is given by  R=R_{0}{A}^{1/3} where,  R  is the radius of the nucleus             A  is the Mass number of the nucleus           and  R_{0}  is a constant  =1.2	imes 10^{-15} Dividing both sides by  R_{0} , we get: \dfrac{R}{R_{0}} = A^{1/3} Taking logarithm of both sides: ln \dfrac{R}{R_{0}} = \dfrac{1}{3}ln A This is in the form of a straight line,  Y=mX , where  Y= ln(\dfrac{R}{R_{0}})   and   X=ln A

Question 3

(i) What characteristic property of nuclear force explains the constancy of binding energy per nucleon

(B E / A)

in the range of mass number

^{'} A^{'}

lying

30 < A < 170

?
(ii) Show that the density of nucleus over a wide range of nuclei is constant-independent of mass number

A

.

Accepted Answer

(i)The approximate constancy of  \dfrac{BE}{A}  over most of the range is saturation property of nuclear force.In heavy nuclei, nuclear size > range of the nuclear force. So a nuclear sense approximately a constant number of neighbors and thus the nuclear  \dfrac{BE}{A}  levels off at high A. This saturation of the nuclear force. (ii) The mass of the nucleus of the atom of the mass number AA=  A a.m.u.   = A  	imes 1.660565 	imes 10^{-27} kg The volume of the nucleus is   \dfrac{4 \pi r^{3}}{3}= \dfrac {4 \pi}{3} (R_0 A^{-3})^{3}= \dfrac {4 \pi R_0^{3} A}{3}  	herefore Volume = \dfrac {4 \pi}{3}(1.1 	imes 10^{-15})^{3} 	imes A m^{3} Now the density of the nucleus=   \dfrac {m}{ V} ; where m is mass and V is the volume of the nucleus. ho = \dfrac {A 	imes 1.660565 	imes 10^{-27}}{\dfrac{4 \pi}{3}(1.1 	imes 10^{-15})^{3} 	imes A}= 2.97 	imes 10^{17} {Kg} m^{-3} Thus, we can see the density of the nuclei is independent of mass number and it is constant for all nuclei .

Question 4

Calculate the binding energy per nucleon for Beryllium  _{4}{Be}^{9} , its mass being  9.012 \ u . The masses of proton and neutron are  1.008  u  and  1.009  u . (Take  1  u = 931.5  MeV )

Accepted Answer

First, let us find the mass defect. The Beryllium nucleus contains 4 protons and 5 neutrons.Mass of 4 protons  = 4 	imes 1.008 = 4.032  u ;  Mass of 5 neutrons  = 5 	imes 1.009  u = 5.045  u Total mass of protons (4) and neutrons (5)  = 4.032 + 5.045 = 9.077  u Mass defect  = 9.077 - 9.012 = 0.065  u The mass defect converted into equivalent energy gives binding energy. 1  u = 931.5  MeV 	herefore   0.065  u = 0.065 	imes 931.5  MeV = 60.5475  MeV Binding energy per nucleon  = \displaystyle\frac{Binding   energy}{Number   of   nucleons} = \displaystyle\frac{60.5475}{9} = 6.7275  MeV

Question 5

Consider the nuclear fission,  Ne^{20} \rightarrow 2He^4 + C^{12} Given that the binding energy/nucleon of  Ne^{20} ,  He^4  and  C^{12}  are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement:

Accepted Answer

Energy absorbed; \Delta Q = 	ext{Binding energy of products - Binding energy of reactants} 	ext{Binding energy of products} = 2 	imes (	ext{B.E of  }  He^4 ) + (	ext{B.E of  }  C^{12} ) = 2 (4 	imes 7.07 ) + (12 	imes 7.86) =150.88 \ MeV 	ext{Binding energy of Reactants}  =  20 	imes 8.03 = 160.6 \ MeV \Delta Q = 	ext{Binding energy of products - Binding energy of reactants} = 150.88 - 160.6 = -9.72 \ MeV 	ext{ Hence, 9.72 MeV energy will be released}

Question 6

A nucleus with mass number  220  initially at rest emits an  \alpha -particle. lf the  \mathrm{Q}  value of the reaction is  5.5    \mathrm{M}\mathrm{e}\mathrm{V} , calculate the kinetic energy of the  \alpha -particle.

Accepted Answer

Let the velocity of alpha particle be  v .Mass of the alpha nucleus is 4amu.From conservation of momentum, 216(v_{atom})=4v \implies v_{atom}=\dfrac{v}{54} Total energy  =Q=5.5MeV=\dfrac{1}{2}(4)v^2+\dfrac{1}{2}(216)(\dfrac{v}{54})^2 Solving gives KE of alpha particle  =\dfrac{1}{2}(4)v^2=5.4MeV

Question 7

Find the energy released when two  _1 H^2  nuclei fuse together to form a single  _2 He^4  nucleus. Given, the binding energy per nucleon of  _1H^2  and  _2He^4  are  1.1  MeV and  7.0  MeV respectively.

Accepted Answer

Binding every per nucleon in  _1H^2 (Deuterium) = 1.1MeV Binding energy per nucleon in  _2He^4(Helium) = 7.0MeV Fusion equation is _1H^2 + _1H^2 	o 2He^4 + \Delta E Total binding energy of reactant (Deuterium)  = 2 [ 2	imes 1.7] = 2[2.2]MeV =4.4MeV     ...(i)Total binding energy of product is = 4[7.0] = 28MeV =28 MeV      ...(ii)Hence, energy released in fusion is = 28 - [4.4] =23.6MeV Hence  23.6MeV  energy is released when two deuterium fuses to form a helium nucleus.

Question 8

20 % of a radioactive element disintegrates in  1hr .The percentage of the radioactive element disintegrated in  2hrs  will be

Accepted Answer

Let initially the number of radioactive particles be  100 .Using the formula: N = N_0 e^{-\lambda t}   where  \lambda = \dfrac{2.303}{t} log(\dfrac{N_0}{N})   \lambda = \dfrac{2.303}{1} log(\dfrac{100}{80}) \Rightarrow  \lambda = 0.223  hrs^{-1} Radioactive material left in  2\ hrs    = N_0 e^{-\lambda t}                                                           = 100 e^{-0.223	imes2}                                                           =63.99 \approx 64% Radioactive material decayed in 2 hours   = (100-64)%                                                                        = 36  %

Question 9

In a Uranium ore the ratio of  U^{238}  nuclei to  Pb^{206}  nuclei is   \eta = 2.8 . Evaluate the age of the ores, assuming all the lead  Pb^{206}   to be a final decay product of the uranium series. The half life of  U^{238}   nuclei is  4.5 	imes 10^{9}   years.

Accepted Answer

What this implies is that in the time since the ore was formed,   \dfrac{\eta}{1 + \eta} U^{238}   nuclei have remained undecayed. Thus                           \dfrac{\eta}{1 + \eta} = e^{-t 	imes \dfrac{ln\ 2}{T_{1/2}}}   Taking logarithm both side , we get or                      t = T_{\dfrac{1}{2}} \dfrac{ln \dfrac{1 + \eta}{\eta}}{ln\ 2}   Substituting   T_{\dfrac{1}{2}} = 4.5 	imes 10^{9}   years ,   \eta = 2.8  we get            t = 1.98 	imes 10^{9}   years

Nuclei