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Nuclei Test - 54

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Nuclei Test - 54
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  • Question 1
    1 / -0
    For mass defect of 0.3%0.3\%, the binding energy of 1kg1kg material is :
    Solution

    Δm=0.3% of 1 kg\Delta m=0.3 \% \text { of } 1 \mathrm{~kg} \\
    Δm=0.3100\Delta m=\frac{0.3}{100} \\
    Δm=3×103 kg\Delta m=3 \times 10^{-3} \mathrm{~kg} \\
    E=(Δm)c2E=(\Delta m) c^{2} \\
    E=3×103×(3×108)2E=3 \times 10^{-3} \times\left(3 \times 10^{8}\right)^{2} \\
    E=3×9×103×1016 JE=3 \times 9 \times 10^{-3} \times 10^{16} \mathrm{~J} \\
    [B]E=2.7×1014 J{[B] E=2.7 \times 10^{14} \mathrm{~J}}

  • Question 2
    1 / -0
    Write the S.I. unit of activity.
    Solution
    S.I.S.I. Unit of activity : It is the count rate of radioactivity.
    It's S.IS.I Unit is decay per second =1 Bq=1\ Bq
    where Bq=Becquerel (1 Bq=1 decay persecond)Bq=Becquerel  (\because 1\ Bq=1\ decay\ per second)
  • Question 3
    1 / -0
    The wavelength of  KαK _ { \alpha }  line for an element of atomic number  4343  is  λ.\lambda.  Then the wavelength of  KαK _ { \alpha }  line for an element of atomic number  2929  is
    Solution
    As we know,
    λ1(Z1)2λ2λ1=(Z11Z21)2\lambda \propto \dfrac{1}{(Z-1)^2} \Rightarrow \dfrac{\lambda_2}{\lambda_1}=\left(\dfrac{Z_1-1}{Z_2-1}\right)^2
    λ2λ=(431291)2=(4228)2λ2=94λ\Rightarrow \dfrac{\lambda_2}{\lambda}=\left(\dfrac{43-1}{29-1}\right)^2=\left(\dfrac{42}{28}\right)^2 \Rightarrow \lambda_2=\dfrac{9}{4} \lambda
  • Question 4
    1 / -0
    The nuclear radius is given by R=r0A1/3R = r_0A^{1/3} , where r0r_0 is constant and A is the atomic mass number.
    Then :
    Solution
    C|C|
    R=r0A1/3R=r_0A^{1/3}
    R3=r03AR^3=r^3_0A
    Density=MassVolume=Amp43πR3=Amp43πr03A=3mp4πr03Density=\dfrac{Mass}{Volume}=\dfrac{Am_p}{\dfrac{4}{3}\pi R^3}=\dfrac{Am_p}{\dfrac{4}{3}\pi r^3_0A}=\dfrac{3m_p}{4\pi r^3_0}
    Hence mass density does not depend on mass (A) or atomic number (z)
  • Question 5
    1 / -0
    Protons of energy 22eV and 2.52.5eV successively illuminate a metal whose work function is 0.50.5eV. The ratio of maximum speed of emitted electron is _______.
    Solution
    K.Emax=hvϕK.E_{max}=hv-\phi
    12mv2=hvϕ\dfrac{1}{2}mv^2=hv-\phi
    v12v22=20.52.50.5=1.52=34\dfrac{v^2_1}{v^2_2}=\dfrac{2-0.5}{2.5-0.5}=\dfrac{1.5}{2}=\dfrac{3}{4}
    v1v2=32\dfrac{v_1}{v_2}=\dfrac{\sqrt{3}}{2}.
  • Question 6
    1 / -0
    The notation for an isotope of sodium 1123Na^{23}_{11}Na. Which row gives the composition of a neutral atom of this isotope of sodium?

    number of protonsnumber of neutronsnumber of electrons
    A111112121111
    B111112121212
    C111123231111
    D121211111212
    Solution
    In 1123Na^{23}_{11}Na 
    Atomic No. , A=11A= 11
    Mass No. ,M=23M= 23

    Number of protons= A=11A= 11
    As it its not ionized 
    Number of electrons =A=11=A=11

    Number of neutrons,=MA=2311=12=M-A=23-11=12
  • Question 7
    1 / -0
    Molecular mass of dry air is ________________.
    Solution
    The molecular mass of dry ice is greater than moist air.
    As dry air consists of nitrogen and oxygen while moist air contains water vapour which less mass than that of oxygen and nitrogen making the mass of moist air lower.

    M.MM.M of dry air 28.9 gm\simeq 28.9\ gm 

    M.MM.M of moist air <28.9 gm< 28.9\ gm
  • Question 8
    1 / -0
    Which diagram represents a nucleus of 13H^3_1H?
    Solution
    Here 
    n== a neutron
    p== a proton.

    In 13H^3_1 H
    As its Atomic No. is 11. So the number of protons is 11
    And its  Mass no is 33

    No. of neutrons = Atomic No. - Mass No. =31=2=3-1=2

    So follwing diagram is correct structure of 13H^3_1H

  • Question 9
    1 / -0
    When the atomic number AA of a nucleus increases :
    Solution

  • Question 10
    1 / -0
    n=p+en=p+e, n=n= neutron, p=p= proton, e=e= electron 
    The decay equation is 
    Solution

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