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Nuclei Test - 64

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Nuclei Test - 64
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  • Question 1
    1 / -0
    The radio-isotope used in agriculture is:
    Solution
    The radio-isotope used in agriculture is $$_{15}P^{32}$$. It is used to trace the assimilation of phosphorus by plants.
    $$_{15}P^{32}$$ is also used in treatment of leukemia. $$_{11}Na^{24}$$ is used to detect tumors and blood clots. $$_{15}P^{30} (\beta)$$ is used in treatment of skin diseases.
  • Question 2
    1 / -0
    A nucleus $$^{A}_{Z}X$$ has mass represented by $$m(A, Z)$$. If $$m_p$$ and $$m_n$$ denote the mass of proton and neutron respectively and BE the binding energy(in MeV) then.
    Solution
    In the case of formation of nucleus the evolution of energy equals to the binding energy of the nucleus takes place due to disappearance of a fraction of total mass. If the quantity of mass disappearing is $$\Delta$$m, then the binding energy is
    $$BE=\Delta mc^2$$
    From the above discussion, it is clear that the mass of nucleus must be less than the sum of the masses of the constituent neutrons and protons. We can then write
    $$\Delta m=Zm_p-Nm_n-m(A, Z)$$
    where $$m(A, Z)$$ is the mass of the atom of mass number A atomic number Z. Hence, the binding energy of nucleus is
    $$BE=[Zm_p+Nm_n-m(A, Z)]c^2$$
    $$BE=[Zm_p+(A-Z)m_n-m(A, Z)]c^2$$.
  • Question 3
    1 / -0
    Outside nucleus
    Solution

  • Question 4
    1 / -0
    Find Binding energy of an $$\alpha-$$particle in $$MeV$$?
    $$[m_{proton}=1.007825\ amu, m_{neutron}=1.008665\ amu, m_{helium}=4.002800\ amu]$$
    Solution
    $$\begin{array}{rl}\text { Given }- & * m_{\text {proton }}=1.007825\text { amu } \\&* m _{\text { neutron }}=1.008665 \text { amu } \\& * m_{\text {Helium }}=4.002800 \text { amu. }\end{array}$$ 
    $$\text{For the reaction below-}$$
    $$2 \cdot{ }_{1}^{1} \mathrm{H}+2 \cdot_{ 0}^{1} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}$$

    $$\begin{array}{l}\text { We get ,Binding energy of }_{2}^{4}\mathrm{He}  \\\Rightarrow B E=\Delta m \cdot c^{2}=\left[m_{\text {Helium }}-2 \mathrm{~m}_{\text {Proton }}-2\mathrm{~m}_{\text {neutron }}\right] c^{2}\\\qquad\begin{aligned}B E=&(0.03018) \text { amu } \times c^{2}=0.03018 \times 931\mathrm{MeV}\\&=28.097\mathrm{MeV}\end{aligned}\end{array}$$
  • Question 5
    1 / -0
    A neutron of kinetic energy $$65\ eV$$ collides inelastically with a singly ionized helium atom at rest. It is scattered at an angle of $$90^{\circ}$$ with respect to its original direction.
    Find the allowed values of the energy of the neutron and that of the atom after the collision.
    [Given : Mass of $$He\ atom = 4\times$$ (mass of neutron) Ionization energy of $$H$$ atom $$= 13. 6\ eV$$]
    Solution
    solution:
    Applying conservation of linear momentum in horizontal direction
    (Initial Momentum )x = (Final Momentum)x
    $$(P1)_x = (Pf)_x$$
    $$\Rightarrow \sqrt{2}Km = \sqrt{2}(4m)K_1 cos θ …(i)$$
    Now applying conservation of linear momentum in Y – direction
    $$(Pi)_y = (Pf)_y$$
    $$= \sqrt2K_2 m - \sqrt 2 (4m) K_1 sin θ$$
    $$\rightarrow \sqrt 2 K_2 m = \sqrt 2(4m)K_1 sin θ …. (ii)$$
    Squaring and adding (i) and (ii)
    $$2Km + 2Km2 m = 2 (4m) K_1 + 2 (4m)K_1$$
    $$K_1 + K_2 = 4K_1 ⇒ K = 4 K_1 – K_2 ⇒ 4K_1 – K_2 = 65 …(iii)$$
    When collision takes place, the electron gains energy and jumps to higher orbit.
    Applying energy conservation
    $$K = K_1 + K_2 + \Delta E$$
    $$65 = K_1 + K_2 + \Delta E$$
    Possible value of $$\Delta E$$ for He+
    Case (1)
    $$\Delta E_1 = - 13. 6 – (- 54.4 ) = 40.8 eV$$
    $$K_1 + K_2 = 24.2 eV$$ from (4)
    Solving with (3), we get
    $$K_2 = 6.36 eV ; K_1 = 17 .84 eV$$
    Case (2)
    $$\Delta E_2 = - 6.04 – (- 54 .4) = 48 .36 eV$$
     $$K_1 + K_2 = 16.64 eV$$ from (4)
    Solving with (3), we get $$K_2 = 0.312 eV; K_1 = 16. 328 eV$$

    Hence the correct option: A
  • Question 6
    1 / -0
    Nuclei of a radioactive element $$A$$ are being produced at a constant rate $$\alpha$$. The element has a decay constant $$\lambda$$. At $$t =0$$, there are $$N_{0}$$ nuclei of the element.
    If $$\alpha = 2N_{0}\lambda$$, calculate the number of nuclei of $$A$$ after one half life of $$A$$, and also the limiting value of $$N$$ as $$t\rightarrow \infty$$.
    Solution
    $$\textbf{Solution}:$$

    If $$\alpha=2 \lambda N_{0}$$, t = half life $$= \dfrac{ln\left(2\right)}{\lambda}$$ 
    $$\therefore N=\dfrac{1}{\lambda}\left[2\,\lambda N_{0}-\left(2\lambda N_{0}-\lambda N_{0}\right)e^{-\lambda t}\right]$$
     
    or $$N=\dfrac{\lambda N_{0}}{\lambda}\left[2-e^{-ln\left(2\right)}\right]$$  [Here $$e^{-ln(2)}=2^{-1}=\dfrac{1}{2}]$$
     
    or $$N=\dfrac{\lambda N_{0}}{\lambda}\left[2-\dfrac{1}{2}\right]=\dfrac{3N_{0}}{2}$$ or
    $$N=\dfrac{3}{2}N_{0}$$ 

    When $$t\rightarrow\infty$$ and $$\alpha=2\lambda N_{0}$$ 

    $$N=\dfrac{\alpha}{\lambda}=\dfrac{2\lambda N_{0}}{\lambda}=2N_{0}$$ or $$N=2N_{0}$$


    $$\textbf{Hence B is the correct option}$$
  • Question 7
    1 / -0
    A system of binary stars of masses $$m_A$$ and $$m_B$$ are moving in circular orbits of radii $$r_A$$ and $$r_B$$, respectively. If $$T_A$$ and $$T_B$$ are the time periods of masses $$m_A$$ and $$m_B$$, respectively then.
    Solution
    1) From the center of mass, we get
    $$M_{1}R_{1}=M_{2}R_{2}$$-------------1

    2) Force acting on both stars i.e. gravitational force and centripetal force are balanced.
    $$\cfrac{GM_{1}M_{2}}{R^{2}}=\cfrac{M_{1}V_{1}^{2}}{R_{1}}=\cfrac{M_{1}V_{1}^{2}}{R_{1}}$$-------------2
    As we know, $$V_{1}=\cfrac{2\pi R_{1}}{T_{1}}$$ & $$V_{2}=\cfrac{2\pi R_{2}}{T_{2}}$$
    Substituting Equation 3 in Equation 2, we get
    $$\cfrac{M_{1}R_{1}}{(T_{1})^{2}}=\cfrac{M_{2}R_{2}}{(T_{2})^{2}}$$------------4
    From Equation 1, we conclude in Equation 4
    $$(T_{1})^{2}=(T_{2})^{2}$$ or $$T_{1}=T_{2}$$

  • Question 8
    1 / -0
    If $${M}_{o}$$ is the mass of an oxygen isotope $$ { _{ 8 }^{  }{ O } }^{ 17 },{ M }_{ P },{ M }_{ N }$$ are the masses of a proton and a neutron respectively, the nuclear binding energy of the isotope is (The speed of light is C)
    Solution
    $$\begin{aligned}\text { Given - }& *\text { Mass of } _8O^{17}\text { isotope }=M_{0} \\& *\text { Mass of neutron }=M_{N} \\& *\text { Mass of proton }=M_{p}\end{aligned}$$


    $$\text{ For the decay of the Isotope we can write-}$$
    $$_8O^{17} \longrightarrow{ }_{8}O^{16}+H^{+}(\text {Neutron) }$$
    $$\text{ For the decay of the Isotope we can write-}$$

    $$\begin{array}{l}B E=\Delta m \cdot c^{2}=\left(M_{\text {Reactant }}-M_{\text {product }}\right) \cdot C^{2} \\B E=\left(M_{0}-M_{\text {product }}\right) \cdot c^{2}\end{array}$$

    $$\begin{array}{l}M \text { product } \Rightarrow \text { we have } 8\text { protons and } 9 \text { neutrons } \\M_{\text {product }}=8 M_{p}+9 M_{N} \\\Rightarrow \quad B E=\left(M_{0}-8 M_{p}-9 M_{N}\right) \cdot c^{2}\end{array}$$



  • Question 9
    1 / -0
    Binding energies of $$ _1H^2 , _2He^4 , _{26}Fe^{56} , $$ and $$_{92}U^{235}$$ nuclie are $$2.22 Me V,28.4 MeV , 492MeV $$ and $$1786MeV$$ respectively which one of the following is more stable?
    Solution

    The binding energy is the energy that is required to break the nucleons apart. The curve of Binding energy per nucleon Vs the Mass number shows the highest binding energy is that of Iron no other is tightly bound. Except the light nuclei having BE.$${\rm{56MeV}}$$.

    Hence the most stable nucleon is $${\rm{2}}{\rm{.22MeV}}$$

  • Question 10
    1 / -0
    Assuming that $$200\ MeV$$ of energy is released per fission of $$_{92}U^{235}$$ atom. Find the number of fission per second ,required to release $$1\ kW$$ power.
    Solution
    $$\begin{array}{l}\text { Given, } \\\text { Energy released per fission }=200 \times 10^{6}\mathrm{eV} \\\text { Power require }=1\times 10^{3} \text { watt } \\\text { Energy in Joule }=200 \times 10^{6} \times 1.6 \times 10^{-19} \mathrm{~J} \\\therefore 1\mathrm{ev}=1.6 \times 10^{-19} \mathrm{~J}\text { oule } \\\qquad=3.2 \times 10^{-11} \text {Joule }\end{array}$$
    $$P=\frac{\text { Energy }}{time}$$
    $$\text { let n : no of fission occur per second }$$
    $$\text { Energy released will be }=n\left(3.2\times 10^{-11} \mathrm{~J}\right)$$
    $$\begin{aligned}10^{3} &=n\left(3.2 \times 10^{-11}\right) \\n &=\frac{1}{3.2} \times 10^{14}\\&=0.3125 \times 10^{14} \\n &=3.125 \times 10^{13}\end{aligned}$$
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