13. NUCLEI (Long Answer Type Question)

Q.1. (i) Find the energy released in the nuclear reaction A = B + C + D the binding energies of A, B, C and E_A, E_B, E_C and E_D.
(ii) Nuceli are being added at the rate N nuceli per second in a sample. Its decay constant is λ At t = 0, the number of nuclei is N₀. Determine the number at time t.

Ans ⇒ (i) Free energy (m_A + m_B – m_C – m_D) C²
           Now, E = bound energy of nucleus A = (Z_Am_p + N_Am_n – m_A)C² (NA = No. of neutrons is A)
.       

             (ii) Number of nucleus added during time dt = ndt and number of disingrated nucleus = -λNdt + dt
             Hence no. of reduced nucleus = -dn = -λNdt + ndt
.           

. 

Q.2. Draw a plot showing the variation of binding energy pernuclear with mass mass number A. Write two important conclusions which you can draw this plot. Explain with the help of this plot, the release in energy in the processes of nuclear fusion and fission.

Ans ⇒ The average energy required to release a nucleon from the nucleus is called binding energy per nucleon.
              Binding energy per nucleon =
.           

Significance of Binding energy per nucleon :
Binding energy per nucleon determines the stability of a nucleus. That is, stability of a nucleus is proportional to the binding energy/nucleon.

           If binding energy per nucleon of a nucleus is less, the nucleus is less stable whereas the nucleus is more stable if its binding energy per nucleon is higher.
          The variation of binding energy per nucleon with mass number (A) is shown in the adjoining figure.

From the graph, it is clear that :

               (i) For nuclei with mass number A < 20 the binding energy per nucleon increases first rapidly and then gradually till it attains a maximum value 8.8 MeV corresponding to lorn, ₂₆Fe⁵⁶ nucleus. Thus, Iron is a stable element.

              (ii) The peaks in the graph for neclei A < 20 show that some light nuclei are more stable than their neighbours. For example, peaks occur at A = 4, 8, 12, 16 and 20. Thus, ₂He⁴, ₄Be⁸, ₆C¹², ₈O¹⁶ and ₁₀Ne²⁰ are most stable than their neighbours.

             (iii) For nuclei having A > 56, binding energy per nucleon gradually decreases. For uranium U²³⁸, one of the heaviest element, the value of B.E./nucleon drops to 7.5 MeV.

          Conclusions : From the above graph, we conclude that.
         (i) The binding energy per nucleon has low value for both the light and heavy nuclei. So they are unstable nuclei.

        (ii) The intermediate nuclei have large value of binding energy per nucleon, so they are more than the neighbours. This shows that these nuclei are more stable.

        (iii) Binding energy per nucleons of light nuclei having mass number A = 4, 8, 12, 16 and 20 is more than the neighbours. This shows that binding energy per nucleon is less for light atoms so energy is released if they combine i.e. when fusion occurs.
                   The graph also shows that heavy nuclei are unstable so they suffer fission and energy is released thereof.

Q.3. Derive the relation connecting half-life period and decay constant of a radioactive isotope.

Ans ⇒ Let at T = 0, the number of nuclei present be N₀. Then the number of nuclei available at t = T, 2T, 3T, etc, is given by
.                         

                  If the time elapsed is not an exact multiple of half life, then we adopt the following method. Let the total number of nuclei present in an element be N. Let dN be the nuclei that will decay in time dt, then dN, is proportional to N and dT i.e.,
                        dN α NaT; dN = -λNdT                          …(i)
                 Here λ is constant, called decay constant. The -ve sign shows that N is decreasing with increase in time. This

       or,       log, N = λt + C                       … (ii)
        Where C is constant of proportionality.
        When t = 0, N, = N₀ hence
        log_eN₀ = C
        Putting the value of C in the above equation.
        We have,
                      log_eN = -λt + log_e N₀
.       

.     

             The above relation shows that the number of remaining nuclei decreases exponentially with time. Graphi plotted between of remaining atoms and time is shown in figure.
Substituting the value of t = 1/λ in equation N = N₀e^-λt, we get
.                               

Thus decay constant or disintegration constant is the reciprocal of the time. During which the original number of atoms (N₀) of a radioactive substance reduces 1/e time the original number of atoms (N₀).

Class 12th physics Long Type question in English