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Physics Part II

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Class 10 Class 12
Two nuclei have mass numbers in the ratio 27:125. What is the ratio of their nuclear radii?

Rato of mass numbers , A1:A2 = 27:125 

                                   A1A2 = 27125

Radius of nucleus is, R = R0 A1/3 

                     R1R2 = A1A21/3         = 271251/3         = 35

Thus, Ratio of radii is 3:5. 


Define nuclear force. Give its two most important characteristics.What is the energy released if all the deuterium atoms in a lake of cross sectional area 2.56 × 105 (km)2 and depth 80 m is used in fusion? Given abundance of H12 = 0.0156% of hydrogen density of water = 103 kg m-3 energy released due to fusion of one atom of H12 = 7.17 MeV. 

Nuclear force is a strong attractive force acting between the nucleons of the atomic nucleus that holds the nucleus together. 

Characteristics of nuclear force : 

i) Nuclear forces are short range forces operating upto a distance which is of the order of a few fermi. 

ii) Nuclear forces are strongest force in nature. It's magnitude is 100 times that of the electrostatic force and 1038 times that of the gravitational force. 


The volume  V of lake is, V = 2.56 ×105 × 106 × 80 m3   = 2.048 × 1013 m3 

 Mass of water, M in lake  = 2.048 x 1013 x 103
                                          = 2.048 x 1016 kg
                                          = 2.048 x 1019 gm 

Now, 18 g of water contains 2 g of hydrogen. 

 Mass of hydrogen atom in lake = 19×2.048×1019g 

 Number of atoms of hydrogen in lake = 
6.023 × 1023 × 2.048 × 10199
                                                               = 1.37 × 1042 

Since abundance of H12 is only 0.0156% of hydrogen atoms, the number of H12 atoms in lake, 

= 1.37 × 1042 × 1.56 × 10-4 = 2.137 × 1038 

Energy released due to fusion of one atom of   H12 = 7.17 MeV 

 Energy released when all H12 atoms present undergo fusion, 

              = 2.137 × 1038 × 7.17 MeV= 1.532 × 1039 MeV.

State the laws of radioactivity.
A radioactive substance has a half-life period of 30 days. Calculate (i) time taken for 34 of original number of atoms to disintegrate and (ii) time taken for 18 of the original number of atoms to remain unchanged.

In any radioactive sample, undergoing α, β or γdecay, the number of nucleii undergoing decay per unit time is directly proportional to the total number of nuclei in the sample. This is known as the radioactive decay law. 

Let, N be the number of nucleii in the sample, 
N is the sample undergoing decay and, 
t  is the time then, 

                            Nt  N 

                        Nt= λN 

where, λ is the decay constant. 


Half -period of radioactive substance = 30 days
Number of atoms disintegrated = 34  N0
Number of atoms left after time t,  N = N0 - 34N0 = 14N0 

Number of half lives in time t days,
                        n = tT = t30

T = Half life time
n = no. of half lives
t = time for disintegrates 

Number of nuclei left after n half lives is given by, 

                      N = N012N 
                    N04 = N012t/30  

            (2)t/30 = 4 = (2)2

               t30 = 2  or  t = 60 days

(ii) Now, using the formula,  N = N012n

                    N08 = N012t/30

               (2)t/30 = 8 = (2)3      
                  t30 = 3
i.e.,                      t = 90 days, is the time taken for 1/8 of the original number of atoms to remain unchanged.



Define the term decay constant of a radioactive nucleus.
Two nuclei P, Q have equal number of atoms at t = 0. Their half lives are 3 hours and 9 hours respectively. Compare their rates of disintegration, after 18 hours from the start.

Decay constant of a radioactive element is the reciprocal of the time during which the number of atoms left in the sample reduces to 1e times the number of atoms in the original sample. 

Given, two nucleii  P and Q. 
P and Q have equal number of atoms at t=0. 
Half life of P = 3 hours. 
Half life of Q = 9 hours. 

and, t = 18 hours

Number of half lives of P in 18 hours  =tT12183=6 

Number of nuclei P left undecayed after 6 half lives is, 
                      N1 = N126 

Number of half lives of Q in 18 hours=tT12 = 189=2 

Number of nuclei of Q left undecayed after 2 half lives is, 

                       N2 = N122 

Ratio of their decay rate is, 

               R1R2 = λ1 N1λ2 N2 = T2 N1T1 N2 

  R = λ N and T = 0.693λ 

               R1R2 = 93×N126N122 = 316 

Hence ratio of disintegration is, R1:R2 = 3 :16

Group the following six nuclides into three pairs of (i) isotones, (ii) isotopes and (iii) isobars:
 612C,   23He,   80198Hg,  13H,  79197Au,  614C
How does the size of a nucleus depend on its mass number? Hence explain why the density of nuclear matter should be independent of the size of the nucleus.  

Classification of nuclides is as follows: 

IsotopesC 612, C614 ; has the same atomic number but, different mass number. 

Isobars : He23, H13 ; has the same mass number but, different atomic number.

Isotones 80198Hg,  79197Au ; has same neutron number but, different proton number. 


Radius of nucleus, R = R0 A1/3                          ...(i) 
where,  R0 is the range of nuclear force (or Nuclear Unit Radius) 

Density is given by, ρ = M43πR3                        ...(ii) 

But,                         M A                               ...(iii) 

From (i) and (iii), we get

R M1/3  R3  M   R3 = kM where,

kproportionality constant                                 ...(iv) 

From (ii) and (iv), we have
                                   ρ = M43πR3= 34πMKM = constant.

Thus, we can see that nuclear density is clearly independent of mass number A. 


An unstable element is produced in nuclear reactor at a constant rate R. If its half-life β--decay is T1/2, how much time, in terms of T1/2, is required to produce 50% of the equilibrium quantity?

We have, 

Rate of increase of element = number of nuclei by reactor1 second-number of nuclei decaying1 second

That is, 
                 dNdt = R - λN   or   dNdt+λN  = R 
The solution to this is the sum of the homogeneous solution, 
              Nh = ce-λt, where c is a constant, and

a particular solution, Nl = Rλ. 

Therefore, the required solution is, 

                       N = Nh+Np = ce-λt+Rλ 

The constant c is obtained from the requirement that the initial number of nuclei be zero, 
                           N(0) = 0 = c+Rλ  
  c = -Rλ
so that,                N = Rλ(1-e-λt) 

The equilibrium value is (t) = R/λ. 

Setting N equal to 1/2 of this value gives, 

                  12Rλ = Rλ(1-e-λt) 

                       e-λt = 12    

           t = ln 2λ = T1/2 

The result is independent of R.