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Atoms Test - 15

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Atoms Test - 15
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  • Question 1
    1 / -0
    On decreasing principal quantum number nn, the value of rr will :
    Solution
    r=E0h2n2RmZe2r= \dfrac { { E }_{ 0 }{ h }^{ 2 }{ n }^{ 2 } }{ R{ mZe }^{ 2 } }

    So, as nn decreases, rr will also decrease.
  • Question 2
    1 / -0
    The incident photon involved in the photo-electric effect experiment (ν>νo\nu > \nu_o)
    Solution
    Answer is A.
    Incident photon involved in the photoelectric effect may completely disappears or comes out with out change in frequency.
    Case I : When frequency of light is less than threshold frequency than no ejection of electrons takes place so, photon comes out with same
    frequency.
    Case II : When frequency of light is more than threshold frequency, then workfunction energy is consumed into the ejection of electrons and rest energy is consumed into kinetic energy of electron. So, photon completely disappears.
  • Question 3
    1 / -0
    A strong argument for particle nature of cathode rays is that they
    Solution
    Since the electrons have a negative charge, they are repelled by the cathode and attracted to the anode. They travel in straight lines through the empty tube. The voltage applied between the electrodes accelerates these low mass particles to high velocities. Cathode rays are invisible, but their presence was first detected in early vacuum tubes when they struck the glass wall of the tube, exciting the atoms of the glass and causing them to emit light, a glow called fluorescence.
    So, the answer is option (A).
  • Question 4
    1 / -0
    The radius of first Bohr orbit in hydrogen atom is r0r_{0} then the radius of first orbit in helium atom will be
    Solution
    Radius of nth orbit in any atom     rn=0.529n2Zr_n = \dfrac{0.529 n^2}{Z}
    For Hydrogen atom,   Z=1Z = 1  and   r1=ror_1 = r_o
    \therefore    ro=0.529r_o = 0.529
    For Helium,     Z=2Z = 2
    Radius of first orbit in helium    r1 =0.5292=ro2r'_1  =\dfrac{0.529}{2} = \dfrac{r_o}{2}
  • Question 5
    1 / -0
    Cathode rays have velocities
    Solution
    Cathode rays travel with a great velocity nearly 910\dfrac {9}{10}th of the speed of light.

    So, the answer is option (C).
  • Question 6
    1 / -0
    In a Rutherford scattering experiment when a projectile of charge Z1Z_1 and mass M1M_1 approaches a target nucleus of charge Z2Z_2 and mass M2M_2, the distance of closest approach is r0r_0. The energy of the projectile is :

    Solution
    Energy =14π0Z1Z2r0=\frac{1}{4\pi\in_0}\frac{Z_1Z_2}{r_0}
  • Question 7
    1 / -0
    The number of different wavelengths that is possible to be observed in the spectrum from a hydrogen sample if the atoms are excited to third excited state is:
    Solution
    The different wavelengths obtained = (nl)(nl+1)=  \left ( n - l \right ) \left ( n - l + 1 \right )
                                                                 =(31)(3) = \left ( 3-1 \right )\left ( 3 \right )
                                                                 =6 = 6
  • Question 8
    1 / -0
    The angular momentum of an electron in hydrogen atom is proportional to
    Solution

    Hint:  For hydrogen atom, force on an electron due to nucleus is given by,

    F=ke2r2F = k\dfrac{{{e^2}}}{{{r^2}}}

    Where,

    kk is constant,

    ee is charge on electron,

    rr is distance of electron from nucleus.


    Step 1: Note all the given values.

    Formula for centripetal force for a particle of mass mm rotating with velocity vv in radius rr is given as,

    F=mv2rF = \dfrac{{m{v^2}}}{r}

    An electron revolving around a nucleus of the hydrogen atom. Force on the electron is given by,

    F=ke2r2F = k\dfrac{{{e^2}}}{{{r^2}}}

    This force acts as centripetal force on electron responsible for rotation.

    Thus,

    ke2r2=mv2rk\dfrac{{{e^2}}}{{{r^2}}} = \dfrac{{m{v^2}}}{r}

    v=ke2mr \Rightarrow v = \sqrt {\dfrac{{k{e^2}}}{{mr}}}


    Step 2: Find an expression for Angular Momentum. 

    Now, Angular momentum of electron is given as,

    L=mvrL = mvr

    L=mrke2mr \Rightarrow L = mr\sqrt {\dfrac{{k{e^2}}}{{mr}}}

    L=mrke2 \Rightarrow L = \sqrt {mrk{e^2}}

    Since m r k are constants.

    Lr \Rightarrow L \propto \sqrt r

    Thus, the angular momentum of an electron in a hydrogen atom is proportional to the square root of the distance of the electron from the nucleus.

  • Question 9
    1 / -0
    Photon is a
    Solution

  • Question 10
    1 / -0
    If the K.E. of a cathode ray beam is 8KeV, then the tube should work at a Potential Difference of
    Solution
    K.E. = PDx e
    8keV=PDe\rightarrow 8keV = PD * e
    8kV=PD\rightarrow 8kV = PD
    So, p.d. = 8kV
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