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Nuclei Test - 52

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Nuclei Test - 52
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Weekly Quiz Competition
  • Question 1
    1 / -0
    Choose correct statement(s)
    Solution

  • Question 2
    1 / -0
    A person needs glasses for driving is suffering from which defect in his vision 
    Solution

    Near-sightedness, also known as short-sightedness and myopia, is an eye disorder where light focuses in front of, instead of on, the retina. This causes distant objects to be blurry while close objects appear normal

  • Question 3
    1 / -0
    If the mass of $$ proton = 1.008 a.m.u. $$ and mass of $$neutron =1.009 a.m. u.$$, then binding energy per nucleon for $$ _4 Be^9$$ $$(mass = 9.012 a mu )$$ would be:
    Solution
    mass defect is $$\Delta m=4m_p+(9-4)m_n-m_{Be}=(4\times 1.008+5\times 1.009-9.012)amu=0.065amu$$

    so Binding energy of $$_4Be^9$$ will be $$\Delta m \times 931.5Mev=.065\times 931.5Mev=60.5475Mev$$
    or the binding energy per nucleon will be $$\dfrac{BE}{nucleon}=\dfrac{60.5475}{9}=6.7275Mev$$
  • Question 4
    1 / -0
    $$AB_2$$ and $$A_2B_2$$ are two compounds of the elements A and B. 0.25 mole of each of these compounds weights 9 g and 16 g respectively. Find the atomic masses of A and B. 
    Solution

  • Question 5
    1 / -0
    The binding energy per nucleon for the parent nucleus is $$E_1$$ and that for the daughter nuclei is $$E_2$$. Then :
    Solution

  • Question 6
    1 / -0
    The binding energies of the nuclei A and B are $$E _ { a }$$  and $$E _ { b }$$  respectively. Three nuclei of the element B fuse to give one nucleus of element 'A' and an energy 'Q' is released. Then $$E _ { a } , E _ { b } , \underline { Q }$$ are related 
    Solution
    The binding energy of reactants  is $$Q_r=3E_b$$ 
    The binding energy of product  is $$Q_p=E_a$$ 
    the enrgy released will be $$Q=Q_r-Q_p=3E_b-E_a$$
  • Question 7
    1 / -0
    A nuclear transformation is denoted by $$ X(n, \alpha) \  ^7_3 Li $$, then the element X will be :-
    Solution
    Since Nuclear Transformtion is given as 
    $$^A_ZX+^1_0n\rightarrow ^4_2He+^7_3Li$$
    On Applying conservation of mass no. and charge no.
    $$A+1=4+7$$   (mass No. Conservation)
    $$Z+0=2+3$$   (charge No. Conservation)
    $$\Rightarrow A=10$$ and $$Z=5$$
    $$\therefore$$ Element will be   $$^{10}_5B$$
  • Question 8
    1 / -0
    According to binding energy per nucleon versus mass number curve, which is not correct?
    Solution
    According to binding energy per nucleon versus mass number curve-

    Two light nuclei fuse to form medium sized nuclei.
  • Question 9
    1 / -0
    In a hydrogen atom, the binding energy of the electron in the nth state is $$ E_n $$,then the frequency of revolution of the electron in the nth orbit is:
    Solution
    We know that, Kinetic energy=Binding energy
    $$\cfrac { 1 }{ 2 } m{ v }^{ 2 }={ E }_{ n }----(i)\\ mvr=\cfrac { nh }{ 2\pi  } ----(ii)$$
    Dividing (i) by (ii),
    $$\cfrac { v }{ 2\pi r } =\cfrac { 2{ E }_{ n } }{ nh } $$
    Therefore, the frequency of revolution of the electron in the nth orbit is$$=\cfrac { 2{ E }_{ n } }{ nh } $$
  • Question 10
    1 / -0
    The speed of daughter nuclei is
    Solution

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