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Nuclei Test - 11

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Nuclei Test - 11
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  • Question 1
    1 / -0
    $$4_{1}H^{1} \rightarrow  _{2}He^{4}+2e^{+}  + 26 MeV$$
    The above reaction represents 
    Solution
    Overall reaction of fusion is given below
    $$4   _1^1H   \rightarrow   _2^4He   +   2(_1^0e)   +   26MeV$$

    Reaction of fission is
    $$_{92}^{235}U + _0^1n  \rightarrow  _{56}^{141}Ba + _{36}^{92}Kr + 3 _0^1n + Energy$$

    $$\beta $$- decay equation is
    $$_{Z}^{A}X \xrightarrow [ ]{minus   e^-}    _{Z+1}^AY$$

    $$\Upsilon$$ - decay equation is
    $$_Z^AX   \rightarrow    -Z^AX$$
  • Question 2
    1 / -0
    The $$\beta$$-decay process, discovered around 1900, is basically the decay of a neutron $$(n)$$. In the laboratory, a proton $$(p)$$ and an electron $$(e^{-})$$ are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process, i.e. $$n\rightarrow p+e^{-}+\overline{v}_{e}$$, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino $$(\overline{v}_{e})$$ to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is $$0.8\times 10^{6}eV$$. The kinetic energy carried by the proton is only the recoil energy.
    If the anti-neutrino had a mass of 3 $$eV/c^{2}$$ (where $$c$$ is the speed of light) instead of zero mass, what should be the range of the kinetic energy, $$K$$, of the electron?
    Solution
    $$K$$ can't be equal to $$0.8 \times 10^6 eV$$ as anti-neutrino must have some energy
  • Question 3
    1 / -0

    Directions For Questions


    Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, $$^{2}_{1}\mathrm{H}$$ known as deuteron and denoted by $$\mathrm{D}$$ can be thought of as a candidate for fusion reactor. The $$\mathrm{D}-\mathrm{D}$$ reaction is $$^{2}_{1}\mathrm{H}+_{1}^{2}\mathrm{H}\rightarrow_{2}^{3} He +\mathrm{n}+$$ energy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of  $$^{2}_{1}\mathrm{H}$$ 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 $$k=8.6 \times 10^{-5} eV/K; \displaystyle \frac{\mathrm{e}^{2}}{4\pi\epsilon_{0}} =1.44\times 10^{9}\mathrm{e}\mathrm{V}\mathrm{m}$$.

    ...view full instructions

    In the core of nuclear fusion reactor, the gas becomes plasma because of
    Solution
    The gas becomes plasma because of the high temperature. The kinetic energy of the particles are directly proportional to absolute temperature, the high temperature causes the particles to turn into plasma (from gas).
    Hence option D is correct.
  • Question 4
    1 / -0

    Directions For Questions


    Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, $$^{2}_{1}\mathrm{H}$$ known as deuteron and denoted by $$\mathrm{D}$$ can be thought of as a candidate for fusion reactor. The $$\mathrm{D}-\mathrm{D}$$ reaction is $$^{2}_{1}\mathrm{H}+_{1}^{2}\mathrm{H}\rightarrow_{2}^{3} He +\mathrm{n}+$$ energy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of  $$^{2}_{1}\mathrm{H}$$ 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 $$k=8.6 \times 10^{-5} eV/K; \displaystyle \frac{\mathrm{e}^{2}}{4\pi\epsilon_{0}} =1.44\times 10^{9}\mathrm{e}\mathrm{V}\mathrm{m}$$.

    ...view full instructions

    Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each with kinetic energy $$1.5 kT$$, when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature $$T$$ required for them to reach a separation of $$4 \times 10^{-15}$$ m is in the range
    Solution
    Conservation of energy:
    $$ KE  = PE$$
    $$ 2 (1.5kT) = \cfrac{e^2}{4 \pi \epsilon _o d}$$
    $$ 3kT =  \cfrac{(1.6 \times 10^{-19})^2}{4 \pi (8.84 \times 10^{-12} (4 \times 10^{-15}}$$

    $$ T =1.4 \times 10^9 K$$
  • Question 5
    1 / -0
    The nucleus finally formed in fusion of protons in proton-proton cycle is that of :
    Solution
    Overall reaction in proton-proton cycle
    $$4   _1^1H   \rightarrow   _2^4He     +   2(_1^0e)   +   Q$$
    (Proton)  (Helium nucleus)  (Positron)   (energy)
    So, finally Helium nucleus is formed.
  • Question 6
    1 / -0

    Directions For Questions

    The mass of nucleus $$^{A}_{Z}X$$ is less than the sum of the masses of ($$\mathrm{A}-\mathrm{Z}$$) number of neutrons and $$\mathrm{Z}$$ number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. $$\mathrm{A}$$ heavy nucleus of mass $$\mathrm{M}$$ can break into two light nuclei of mass $$\mathrm{m}_{1}$$ and $$\mathrm{m}_{2}$$ only if $$(\mathrm{m}_{1}+\mathrm{m}_{2})<\mathrm{M}.$$ Also two light nuclei of masses $$\mathrm{m}_{3}$$ and $$\mathrm{m}_{4}$$ can undergo complete fusion and form a heavy nucleus of mass $$\mathrm{M}'$$ only if $$(\mathrm{m}_{3} +\mathrm{m}_{4})>\mathrm{M}'$$. The masses of some neutral atoms are given in the table above:

    ...view full instructions

    The kinetic energy (in keV) of the alpha particle, when the nucleus $$^{210}_{84}Po$$ at rest undergoes alpha decay, is

    Solution
    $$^{210}_{84}Po\rightarrow_{2}^{4}\mathrm{H}\mathrm{e}+_{82}^{206}$$ Pb

    $$\mathrm{Q}=(209.982876-4.002603-205.97455)\mathrm{C}^{2}$$
        $$=5.422\mathrm{M}\mathrm{e}\mathrm{V}$$

    From conservation of momentum:
    $$\sqrt{2\mathrm{K}_{1}(4)}=\sqrt{2\mathrm{K}_{2}(206)}$$
    $$4\mathrm{K}_{1}=206\mathrm{K}_{2}$$
    $$\displaystyle \mathrm{K}_{1}=\frac{103}{2}\mathrm{K}_{2}$$
    $$\mathrm{K}_{1}+\mathrm{K}_{2}=5.422$$
    $$\displaystyle \mathrm{K}_{1}+\frac{2}{103}\mathrm{K}_{1}=5.422$$
    $$\Rightarrow  \displaystyle \frac{105}{103}\mathrm{K}_{1}=5.422$$
    $$\mathrm{K}_{1}=5.319\mathrm{M}\mathrm{e}\mathrm{V}=5319\mathrm{K}\mathrm{e}\mathrm{V}$$
  • Question 7
    1 / -0
    The source of stellar energy is 
    Solution
    Stellar and solar energy is due to fusion reactions.
    So, source of stellar energy is Nuclear fusion.
  • Question 8
    1 / -0
    After losing two electrons, an atom of Helium becomes equivalent to
    Solution
    $$ _{2}^{4}He \rightarrow  _{2}^{4}He^{+2} + 2e^{-}$$
    $$_{2}^{4}He^{+2}$$ is alpha particle. Because it has charge equal to  +2e and mass is four times the mass of one proton. 
  • Question 9
    1 / -0
    In the options given below, let $$E$$ denote the rest mass energy of a nucleus and $$'n\ '$$ a neutron. The correct option is 
    Solution
    A bound nucleus has lost the binding energy to become stable and has gained more rest mass energy.
    Rest mass energy of U will be greater than the rest mass energy of the nucleus in which it breaks.
  • Question 10
    1 / -0
    A nucleus with mass number $$220$$ initially at rest emits an $$\alpha$$-particle. lf the $$\mathrm{Q}$$ value of the reaction is $$5.5 $$ $$\mathrm{M}\mathrm{e}\mathrm{V}$$, calculate the kinetic energy of the $$\alpha$$-particle.
    Solution
    Let the velocity of alpha particle be $$v$$.
    Mass of the alpha nucleus is 4amu.
    From conservation of momentum,
    $$216(v_{atom})=4v$$
    $$\implies v_{atom}=\dfrac{v}{54}$$
    Total energy $$=Q=5.5MeV=\dfrac{1}{2}(4)v^2+\dfrac{1}{2}(216)(\dfrac{v}{54})^2$$
    Solving gives KE of alpha particle $$=\dfrac{1}{2}(4)v^2=5.4MeV$$
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