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NCERT Exemplar · Class 12 Physics Nuclei

26 questions

Multiple Choice Questions (Type II) 8–10 (part 2 of 5)

  1. Exercise 8

    Fusion processes, like combining two deuterons to form a He nucleus are impossible at ordinary temperatures and pressure. The reasons for this can be traced to the fact:
    (a)
    nuclear forces have short range.
    (b)
    nuclei are positively charged.
    (c)
    the original nuclei must be completely ionized before fusion can take place.
    (d)
    the original nuclei must first break up before combining with each other.
    NCERT’s answer
    (a), (b)
    (a), (b) Deuterons repel each other, and the attractive nuclear force acts only at about \(\displaystyle 10^{-15}\ \mathrm{m}\), so they must be driven that close against the Coulomb barrier.\[U=\frac{1}{4\pi\varepsilon_0}\frac{e^{2}}{r}\approx\frac{1.44\ \mathrm{MeV\,fm}}{3\ \mathrm{fm}}\approx 0.5\ \mathrm{MeV} \] \[k_BT\,(300\ \mathrm{K})=8.6\times10^{-5}\ \mathrm{eV\,K^{-1}}\times 300\ \mathrm{K}\approx 0.026\ \mathrm{eV}\ll U \] \[T\uparrow\ \Rightarrow\ KE_{avg}\uparrow\ \Rightarrow\ \text{barrier crossed} \]

  2. Exercise 9

    Samples of two radioactive nuclides A and B are taken. λA\displaystyle \lambda_{\mathrm{A}} and λB\displaystyle \lambda_{\mathrm{B}} are the disintegration constants of A and B respectively. In which of the following cases, the two samples can simultaneously have the same decay rate at any time?
    (a)
    Initial rate of decay of A is twice the initial rate of decay of B and λA=λB\displaystyle \lambda_{\mathrm{A}}=\lambda_{\mathrm{B}}.
    (b)
    Initial rate of decay of A is twice the initial rate of decay of B and λA>λB\displaystyle \lambda_{\mathrm{A}}>\lambda_{\mathrm{B}}.
    (c)
    Initial rate of decay of B is twice the initial rate of decay of A and λA>λB\displaystyle \lambda_{\mathrm{A}}>\lambda_{\mathrm{B}}.
    (d)
    Initial rate of decay of B is same as the rate of decay of A at t=2 h\displaystyle t=2 \mathrm{~h} and λB<λA\displaystyle \lambda_{\mathrm{B}}<\lambda_{\mathrm{A}}.
    NCERT’s answer
    (b), (d)
    (b), (d) Equal rates need \(\displaystyle R_A/R_B\) to fall from above $\displaystyle 1$ to exactly $\displaystyle 1$, so A must start higher and decay faster.\[R_A=R_{A0}e^{-\lambda_A t},\qquad R_B=R_{B0}e^{-\lambda_B t} \] \[\frac{R_A}{R_B}=\frac{R_{A0}}{R_{B0}}\,e^{-(\lambda_A-\lambda_B)t}=1 \] \[\text{(a)}\quad \lambda_A=\lambda_B\ \Rightarrow\ \frac{R_A}{R_B}=2\ \text{at every }t \] \[\text{(b)}\quad 2\,e^{-(\lambda_A-\lambda_B)t}=1\ \Rightarrow\ t=\frac{\ln 2}{\lambda_A-\lambda_B}>0 \] \[\text{(c)}\quad \frac{R_A}{R_B}=\tfrac12\,e^{-(\lambda_A-\lambda_B)t}\le\tfrac12\ \text{for all }t\ge 0 \] \[\text{(d)}\quad R_{B0}=R_A(2\ \mathrm{h})=R_{A0}e^{-\lambda_A(2\,\mathrm{h})}\ \Rightarrow\ t=\frac{\lambda_A(2\ \mathrm{h})}{\lambda_A-\lambda_B}>2\ \mathrm{h} \]

  3. Exercise 10

    The variation of decay rate of two radioactive samples A and B with time is shown in Fig. 13.1.
    Which of the following statements are true?
    (a)
    Decay constant of A is greater than that of B, hence A always decays faster than B.
    (b)
    Decay constant of B is greater than that of A but its decay rate is always smaller than that of A.
    (c)
    Decay constant of A is greater than that of B but it does not always decay faster than B.
    (d)
    Decay constant of B is smaller than that of A but still its decay rate becomes equal to that of A at a later instant.
    NCERT_Question_Class12_Physics_Exemplar_Ch13_Q10
    NCERT’s answer
    (c), (d)
    (c), (d) A falls more steeply, so \(\displaystyle \lambda_A>\lambda_B\), yet the curves cross at \(\displaystyle t_1\): A is not always faster, and B's rate later equals A's.\[R=\left|\frac{dN}{dt}\right|=R_0e^{-\lambda t},\qquad \text{steeper fall of A}\ \Rightarrow\ \lambda_A>\lambda_B \] \[t<t_1:\ R_A>R_B,\qquad t=t_1:\ R_A=R_B,\qquad t>t_1:\ R_A<R_B \]