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Nuclei Test - 62

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Nuclei Test - 62
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  • Question 1
    1 / -0

    Directions For Questions

    The compound unstable nucleus $$_{92}^{236}U$$ often decays in accordance with the following reaction:
    $$_{92}^{236}U\rightarrow _{54}^{140}Xe+_{38}^{94}Sr+$$ other particles
    During the reaction, the Uranium nucleus "fissions" (splits) into the two smaller nucleii. The reaction is energetically favorable because the smaller nucleii have higher nuclear binding energy per nucleon (although the lighter nuclei have lower total nuclear binding energies, because they contain fewer nucleons).
    Inside a nucleus, the nucleons (protons and neutrons) attract each other with a "strong nuclear" force. All nucleons exert approximately the same strong nuclear force on each other. This force holds the nucleus together. Importantly, the strong nuclear force becomes important only when the protons and neutrons are very close together at intranuclear distances.

    ...view full instructions

    Which of the following graphs might represent the relationship between atomic number (i.e, "atomic weight") and the total binding energy of the nucleus, for nuclei heavier than $$_{38}^{94}Sr$$?
    Solution
    Binding energy per nucleon decreases, but binding energy increases as atomic number increases. 

  • Question 2
    1 / -0

    Directions For Questions

    All nuclei consist of two types of particles - protons and neutrons. The nuclear force is the strongest force. Stability of nucleus is determined by the neutron-proton ratio or mass defect or binding energy per nucleus or packing fraction. The shape of the nucleus is calculated by quadrupole moment. Spin of a nucleus depends on even or odd mass number. The volume of the nucleus depends on the mass number. The whole mass of the atom (nearly 99%) is centered at the nucleus. The magnetic moment of the nucleus is measured in terms of the nuclear magnetons.

    ...view full instructions

    Binding energy per nucleon is maximum
    Solution
    The above data chart clearly suggests that binding energy per nucleon is maximum for middle order elements.
    This is the reason for formation of stable middle order elements on nuclear fission as well as fusion.

  • Question 3
    1 / -0

    Directions For Questions

    A radionuclide with decay constant $$\lambda$$ is being produced in a nuclear reactor at a rate $$q_0t$$ per second, where $$q_0$$ is a positive constant and $$t$$ is the time. During each decay, $$E_0$$ energy is released. The production of radionuclide starts at time $$t=0$$.

    ...view full instructions

    Instantaneous power developed at time t due to the decay of the radionuclide is
    Solution
    The radioactive nuclide (X) is produced at rate $${ q }_{ 0 }t$$ and decays at the rate $$\lambda \left[ X \right] $$
    Therefore, $$\dfrac { d\left[ X \right]  }{ dt } ={ q }_{ 0 }-\lambda \left[ X \right] $$
    Given, Energy released per decay is $${ E }_{ 0 }$$.
    $$\Rightarrow $$ Instantaneous power at any time t is $$\lambda \left[ X \right] { E }_{ 0 }$$
    Solving, above differential equation, we get:
    $$\left[ X \right] = { q }_{ 0 }\left( \dfrac { t }{ \lambda  } -\dfrac { 1 }{ { \lambda  }^{ 2 } } \left( 1-{ e }^{ -\lambda t } \right)  \right) $$
    Therefore, Instantaneous Power is $${ q }_{ 0 }\left( t-\dfrac { 1 }{ { \lambda  } } \left( 1-{ e }^{ -\lambda t } \right)  \right) { E }_{ 0 }$$
  • Question 4
    1 / -0

    Directions For Questions

    The compound unstable nucleus $$_{92}^{236}U$$ often decays in accordance with the following reaction:
    $$_{92}^{236}U\rightarrow _{54}^{140}Xe+_{38}^{94}Sr+$$ other particles
    During the reaction, the Uranium nucleus "fissions" (splits) into the two smaller nucleii. The reaction is energetically favorable because the smaller nucleii have higher nuclear binding energy per nucleon (although the lighter nuclei have lower total nuclear binding energies, because they contain fewer nucleons).
    Inside a nucleus, the nucleons (protons and neutrons) attract each other with a "strong nuclear" force. All nucleons exert approximately the same strong nuclear force on each other. This force holds the nucleus together. Importantly, the strong nuclear force becomes important only when the protons and neutrons are very close together at intranuclear distances.

    ...view full instructions

    A proton and a neutron are both shot at $$100 ms^{-1}$$ toward a $$_6^{12}C$$ nucleus. Which particle, if either, is more likely to be absorbed by the nucleus?
    Solution
    $$^{12}_6C$$ captures a neutron to form relatively stable element $$^{13}_6C$$.
    $$^{12}_6C$$ $$+^1_0n\rightarrow$$ $$^{13}_6C$$
    This is more stable than the otherwise formed element $$^{12}_7N$$
  • Question 5
    1 / -0

    Directions For Questions

    The compound unstable nucleus $$_{92}^{236}U$$ often decays in accordance with the following reaction:
    $$_{92}^{236}U\rightarrow _{54}^{140}Xe+_{38}^{94}Sr+$$ other particles
    During the reaction, the Uranium nucleus "fissions" (splits) into the two smaller nucleii. The reaction is energetically favorable because the smaller nucleii have higher nuclear binding energy per nucleon (although the lighter nuclei have lower total nuclear binding energies, because they contain fewer nucleons).
    Inside a nucleus, the nucleons (protons and neutrons) attract each other with a "strong nuclear" force. All nucleons exert approximately the same strong nuclear force on each other. This force holds the nucleus together. Importantly, the strong nuclear force becomes important only when the protons and neutrons are very close together at intranuclear distances.

    ...view full instructions

    Why is a $$_2^4He$$ nucleus more stable than a $$_3^4Li$$ nucleus?
    Solution
    Same as electron cloud, in nucleus of helium, protons stay in orbits 1s1s, but with zero angular momentum. Therefore, same as stable fully filled shells, nuclei are stable if they are fully filled

  • Question 6
    1 / -0
    The equation $$4_1^1H^{2}\rightarrow _2^4He^{2+}+2e^{+1}+26\  MeV$$ represents
    Solution
    Atoms combine to form higher compounds by releasing some energy. This process is called Fusion.
  • Question 7
    1 / -0

    Directions For Questions

    Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, $$_1^2H$$, known as deuteron and denoted by D can be thought of as a candidate for fusion reactor. The D-D reaction is $$_1^2H+_1^2H\rightarrow _2^3He+n+energy$$. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of $$_1^2H$$ nuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma. Special techniques are used which confine the plasma for a time $$t_0$$ before the particles fly away from the core. If n is the density (number /volume) of deuterons, the product $$nt_0$$ is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than $$5\times 10^{14} s cm^{-3}$$.
    It may be helpful to use the following: Boltzmann constant, $$k=8.6\times 10^{-5} eV K^{-1}; \dfrac {e^2}{4\pi \epsilon_0}=1.44\times 10^{-9}eV m$$.

    ...view full instructions

    In the core of nuclear fusion reactor, the gas becomes plasma because of
    Solution
    When a substance is heated to high temperatures, all the atoms in it get ionized and start conducting electricity. This state of matter is called plasma.
    Hence, high temperatures inside core helps maintaining plasma.
  • Question 8
    1 / -0

    Directions For Questions

    The mass of nucleus $$_Z^AX$$ is less than the sum of the masses of (A-Z) number of neutrons and Z number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass $$M$$ can break into two light nuclei of mass $$m_1$$ and $$m_2$$ only if $$(m) < M.$$ Also two light nuclei of masses $$m_3$$ and $$m_4$$ can undergo complete fusion and form a heavy nucleus of mass $$M$$ "only if $$(m_3+m_4) > M$$". The masses of some neutral atoms are given in the table below:
    $$_1H^2$$
    1.007825 u
    $$_1^2H$$
    2.014102 u
    $$_1^3H$$
    3.016050 u
    $$_2^4He$$
    4.002603 u
    $$_3^6Li$$
    6.015123 u
    $$_3^7Li$$
    7.016004 u
    $$_{30}^{70}Zn$$
    69.925325 u
    $$_{34}^{82}Se$$
    81.916709 u
    $$_{64}^{152}Gd$$
    151.919803 u
    $$_{82}^{206}Pb$$
    205.974455 u
    $$_{84}^{210}Po$$
    209.982876 u



    ...view full instructions

    The correct statement is
    Solution
    1)  
     $$_{ 3 }^{ 6 }{ Li } \longrightarrow  _{ 2 }^{ 4 }{ He }\quad (\alpha -particle)+ _{ 1 }^{ 2 }{ H }$$
     $$_{ 3 }^{ 6 }{ Li } \longrightarrow  _{ 2 }^{ 4 }{ He }\quad (\alpha -particle)+ _{ 1 }^{ 2 }{ H }\\ { M }_{ _{ 3 }^{ 6 }{ Li } }= 6.015123u\quad \\ { m }_{ _{ 2 }^{ 4 }{ He } }+{ m }_{ _{ 1 }^{ 2 }{ H } }= 4.002603u+ 2.014102u= 6.016705u\\ \Rightarrow  { M }_{ _{ 3 }^{ 6 }{ Li } }> { m }_{ _{ 2 }^{ 4 }{ He } }+{ m }_{ _{ 1 }^{ 2 }{ H } }$$
    Therefore, it can emit an alpha particle.

    2)
    $$_{ 84 }^{ 210 }{ Po } \longrightarrow  _{ 1 }^{ 1 }{ H }+ _{ 1 }^{ 3 }{ H } + _{ 82 }^{ 206 }Pb\\ { M }_{ _{ 84 }^{ 210 }Po }= 209.982876u \\ { m }_{ _{ 82 }^{ 206 }{ Pb } }+{ m }_{ _{ 1 }^{ 3 }{ H } }+{ m }_{ _{ 1 }^{ 1 }{ H } }= 205.974455u+ 3.016050u+ 1.007825u= 209.99833u\\ \Rightarrow _{ 84 }^{ 210 }{ Po }< _{ 1 }^{ 1 }{ H }+ _{ 1 }^{ 3 }{ H }+ _{ 82 }^{ 206 }Pb$$
    Therefore, polonium cannot emit a proton.

    3)
    They cannot undergo fusion as proved in the first example.

    4)
     $$_{ 30 }^{ 70 }{ Zn }+ _{ 34 }^{ 82 }{ Se } \longrightarrow  _{ 64 }^{ 152 }{ Gd }\\ { M }_{ _{ 64 }^{ 152 }{ Gd } }= 151.919803u \\ { m }_{ _{ 30 }^{ 70 }{ Zn } }+{ m }_{ _{ 34 }^{ 82 }{ Se } }= 69.925325u+ 81.916709u= 151.842034u\\ \Rightarrow _{ 30 }^{ 70 }{ Zn }+ _{ 34 }^{ 82 }{ Se }< _{ 64 }^{ 152 }{ Gd }$$
    Therefore, fusion is not possible.
  • Question 9
    1 / -0
    A nucleus with mass number $$220$$ initially at rest emits an $$\alpha \ particle$$. If the $$Q$$ value of the reaction is $$5.5 MeV$$, calculate the kinetic energy of the $$\alpha\ particle$$
    Solution
    $$ M \rightarrow X+ \alpha$$
    $$ E = \cfrac{p^2}{2m}$$
    Let Total energy = $$T$$
    $$ T = E_{\alpha} + E_{X}$$
    $$M = M_{\alpha} + M_{X}$$
    $$M_{X}= 216$$
    Conservation of momentum
    $$ p_{\alpha} = p_{X}$$
    $$ m_{\alpha} E_{alpha} = m_XE_X$$
    $$E_{\alpha} = \cfrac{m_x}{M}T$$
    $$ E_{\alpha} = \cfrac{216}{220} 5.5  = 5.4 MeV$$
  • Question 10
    1 / -0
    Four physical quantities are listed in Column I. Their values are listed in Column II in a random order.
    Column I
    Column II
    p. Thermal energy of air molecules at room temperature
    (i) 0.02 eV
    q. Binding energy of heavy nuclei per nucleon
    (ii) 2 eV
    r. X-ray photon energy
    (iii) 10 keV
    s. Photon energy of visible light
    (iv) 7 MeV
    The correct matching the Column I and Column II is given by
    Solution
    p) : Thermal energy of the air molecules at room temperature                                                                               $$E_{thermal}   \approx    KT  =  1.38 \times 10^{-23}  \times 300     J  =  4.14 \times 10^{-21}   J   =   0.259    eV$$    (approx)

    q) : The graph of binding energy per nucleon vs mass number suggests that the binding energy per nucleon of heavy atom lies between $$7$$ to $$8    MeV$$ 

    r) :  X-Rays have wavelength in the range from $$0.01    nm$$ to $$10    nm$$, $$\implies $$  These have energy range of  $$  \approx     100    eV$$  to    $$100    keV$$  

    s) :  Visible light have wavelength $$\approx   380   nm$$ to $$780     nm$$  which implies that these have energy range $$1.6   eV$$ to $$3.3   eV$$        
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