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Structure of Atom Test - 70

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Structure of Atom Test - 70
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  • Question 1
    1 / -0
    The number of revolutions per second made by an electron in the first Bohr's orbit of hydrogen atom is(Given, r=0.53Aor=0.53\overset{o}{A}).
    Solution
    From Bohr's theory
    mvr=nh2πmvr=\displaystyle\frac{nh}{2\pi}
    For first orbit n=1n=1
    m(rω)r=1×h2πm(r\omega )r=\displaystyle\frac{1\times h}{2\pi}
    mr22πf=h2π\Rightarrow mr^22\pi f=\displaystyle\frac{h}{2\pi}
    f=h4π2mv2\Rightarrow \displaystyle f=\frac{h}{4\pi^2mv^2}
    =6.6×10344×(3.14)2×9.1×1031×(0.33×1010)2=\displaystyle\frac{6.6\times 10^{-34}}{4\times (3.14)^2\times 9.1\times 10^{-31}\times (0.33\times 10^{-10})^2}.
    =6.5×1015Hz6.5\times10^{15} Hz
  • Question 2
    1 / -0
    The highest energy in Balmer series, in the emission spectra of hydrogen is represented by:
    (RH=109737cm1)\left({R}_{H}=109737 {cm}^{-1}\right)
    Solution
    vˉ=1λ=RHZ2[1n121n22]\bar{v}=\dfrac{1}{\lambda}=R_H{Z}^{2}\left[\dfrac{1}{{n}_{1}^{2}} - \dfrac{1}{{n}_{2}^{2}}\right]

    =109737×12[12212]=27434.25cm1=109737 \times {1}^{2} \left[\dfrac{1}{{2}^{2}} - \dfrac{1}{{\infty}^{2}}\right] = 27434.25 {cm}^{-1}

    (series limit will be of highest energy)
  • Question 3
    1 / -0
    The energy of second Bohr orbit of the hydrogen atom is 328kJ mol1-328kJ\ { mol }^{ -1 }. Hence, the energy of fourth Bohr orbit would be:
    Solution
    E 1 n2 \displaystyle E \propto \dfrac {  1 }{  n^2 }

    E4 E2 =( n2n4 )2 \displaystyle \dfrac { E_4  }{ E_2  } = \left(\dfrac {  n_2 }{ n_4  } \right)^2

      E4 =328kJ/mol×(  n2 n4 )2 \displaystyle   { E_4  }={-328 \: kJ/mol } \times \left(\dfrac {   n_2 }{  n_4  } \right)^2

      E4 =328kJ/mol×(  2 4 )2 \displaystyle   { E_4  }={-328 \: kJ/mol } \times \left(\dfrac {   2 }{  4  } \right)^2

    E4 =82kJ/mol \displaystyle { E_4  }={-82 \: kJ/mol }

    Hence, the energy of fourth Bohr orbit would be 82kJ mol1-82kJ\ { mol }^{ -1 }.
  • Question 4
    1 / -0
    Which of the following orbitals has the highest energy?
    Solution
    The orbital energies are compared with the help of n+ln+l rule. The higher value of n+ln+l, higher is the energy of the orbital.

    5d=n+l=75d = n+l =7
    5f=n+l=85f = n+l =8
    6s=n+l=66s = n+l =6
    6p=n+l=76p = n+l =7

    The n+ln+l value is greatest for 5f orbital. 
    Hence it has the highest energy among the given options.
  • Question 5
    1 / -0
    The spectral lines corresponding to the radiation emitted by an electron jumping from higher orbits to first orbit belong to:
    Solution
    The spectral lines corresponding to the radiation emitted by an electron jumping from higher orbits to first orbit belong to Lyman series.
    Lyman series corresponds to transitions to n=1\displaystyle n = 1 level.
    Balmer series corresponds to transitions to n=2\displaystyle n = 2 level.
    Paschen series corresponds to transitions to n=3\displaystyle n = 3 level.
    Pfund series corresponds to transitions to n=4\displaystyle n = 4 level.   
  • Question 6
    1 / -0
    If the shortest wavelength of the spectral line of He+{ He }^{ + } in Lyman series is xx, then longest wavelength of the line in Balmer series of Li2+{ Li }^{ 2+ } is:
    Solution
    Shortest wavelength =n n1={ n }_{ \infty  }\rightarrow { n }_{ 1 }

    1λ min=RHZ2n12\cfrac { 1 }{ { \lambda  }_{ min } } ={ R }_{ H }\cfrac { { Z }^{ 2 } }{ { n }_{ 1 }^{ 2 } }

    for He+,Z=2{He}^{+},Z=2

    1λ min=RH221 2\cfrac { 1 }{ { \lambda  }_{ min } } ={ R }_{ H }\cfrac { { 2 }^{ 2 } }{ { 1 }_{  }^{ 2 } } \quad or RH=14x{ R }_{ H }=\cfrac { 1 }{ 4x }

    for the longest wavelength in Balmer series of Li2+,Z=3,n1=2,n2=3{ Li }^{ 2+ },Z=3,{ n }_{ 1 }=2,{ n }_{ 2 }=3

    1λ max=RHZ2(1n121n22 )\cfrac { 1 }{ { \lambda  }_{ max } } ={ R }_{ H }{ Z }^{ 2 }\left( \cfrac { 1 }{ { n }_{ 1 }^{ 2 } } -\cfrac { 1 }{ { n }_{ 2 }^{ 2 } }  \right)

    1λ max=14x(3)2(1419 )\cfrac { 1 }{ { \lambda  }_{ max } } =\cfrac { 1 }{ 4x } { (3) }^{ 2 }\left( \cfrac { 1 }{ 4 } -\cfrac { 1 }{ 9 }  \right)

    1λ max=94x×536\cfrac { 1 }{ { \lambda  }_{ max } } =\cfrac { 9 }{ 4x } \times \cfrac { 5 }{ 36 }

    λ max=16x5\quad { \lambda  }_{ max }=\cfrac { 16x }{ 5 }
  • Question 7
    1 / -0
    The fast emission line on hydrogen atomic spectrum in the Balmer series appears at:

    [R=R = Rydberg constant]
    Solution
    Rydberg - Riz equation is :

    vˉ=RH[1n121n22]   (Wheren1=2,n2=3)\bar{v}=R_H\left [ \dfrac{1}{n^2_1}-\dfrac{1}{n^2_2} \right ] \,\,\, (Where \, n_1= 2, n_2=3)

    vˉ=RH[9464]=5R36 cm1\therefore \bar{v}=R_H\left [ \dfrac{9-4}{64} \right ] =\dfrac{5R} {36}  cm^{-1}
  • Question 8
    1 / -0
    The magnitude of potential energy of an electron in the second Bohr orbit of a hydrogen atom is:
    [a0a_0 is Bohr radius].
    Solution
    Solution:- (B) h216π2mao2\cfrac{{h}^{2}}{16 {\pi}^{2} m a_o^2}
    According to Bohr's model of an atom, the potential energy of electron having charge e-e is given as-
    P.E.=14πϵ0er(e)P.E. = \cfrac{1}{4 \pi {\epsilon}_{0}} \cfrac{e}{r} \left( -e \right)
    P.E.=14πϵ0e2r\Rightarrow P.E. =- \cfrac{1}{4 \pi {\epsilon}_{0}} \cfrac{{e}^{2}}{r}
    r=ϵ0n2h2πme2\because r = \cfrac{{\epsilon}_{0} {n}^{2} {h}^{2}}{\pi m {e}^{2}}
    P.E.=14πϵ0e2(ϵ0n2h2πme2)\therefore P.E. = \cfrac{1}{4 \pi {\epsilon}_{0}} \cfrac{{e}^{2}}{\left( \cfrac{{\epsilon}_{0} {n}^{2} {h}^{2}}{\pi m {e}^{2}} \right)}
    P.E.=14ϵ02me4n2h2\Rightarrow P.E. = \cfrac{1}{4 {{\epsilon}_{0}}^{2}} \cfrac{m {e}^{4}}{{n}^{2} {h}^{2}}
    Now,
    a0=ϵ0h2πme2{a}_{0} = \cfrac{{\epsilon}_{0} {h}^{2}}{\pi m {e}^{2}}
    e4=ϵ02h4π2m2a02\Rightarrow {e}^{4} = \cfrac{{{\epsilon}_{0}}^{2} {h}^{4}}{{\pi}^{2} {m}^{2} {{a}_{0}}^{2}}
    Therefore,
    P.E.=14ϵ02mn2h2( ϵ02h4π2m2a02)\therefore P.E. = \cfrac{1}{4 {{\epsilon}_{0}}^{2}} \cfrac{m}{{n}^{2} {h}^{2}} \left( \cfrac{{{\epsilon}_{0}}^{2} {h}^{4}}{{\pi}^{2} {m}^{2} {{a}_{0}}^{2}} \right)
    P.E.=h24π2n2ma02\Rightarrow P.E. = \cfrac{{h}^{2}}{4 {\pi}^{2} {n}^{2} m {{a}_{0}}^{2}}
    For second orbit of hydrogen atom,
    n=2n = 2
    P.E.=h216π2ma02\Rightarrow P.E. = \cfrac{{h}^{2}}{16 {\pi}^{2} m {{a}_{0}}^{2}}
    Hence the magnitude of potential energy of an electron in the second Bohr orbit of a hydrogen atom is h216π2ma02\cfrac{{h}^{2}}{16 {\pi}^{2} m {{a}_{0}}^{2}}.
  • Question 9
    1 / -0
    The minimum wavelength absorbed by electron initially present in (n1+1)(n_1+1) shell of Li+2Li^{+2} ion:(during its excitation)
    Solution
    Minimum wave length means maximum energy
    So, n2=n_2= \infty
    1λ =RHZ2(1(n1+1) 21( ) 2 )RH×(3)2(1(n1+1) 2 ) λ=(n1+1) 29R \begin{array}{l} \frac { 1 }{ \lambda  } ={ R_{ H } }{ Z^{ 2 } }\left( { \frac { 1 }{ { { { \left( { { n_{ 1 } }+1 } \right)  }^{ 2 } } } } -\frac { 1 }{ { { { \left( \infty  \right)  }^{ 2 } } } }  } \right) \Rightarrow { R_{ H } }\times { \left( 3 \right) ^{ 2 } }\left( { \frac { 1 }{ { { { \left( { { n_{ 1 } }+1 } \right)  }^{ 2 } } } }  } \right)  \\ \lambda =\frac { { { { \left( { { n_{ 1 } }+1 } \right)  }^{ 2 } } } }{ { 9{ R } } }  \end{array}
    Option B is correct.
  • Question 10
    1 / -0
    Calculate the wavelength of the first line of He+He^{+} ion spectral series whose interval with extreme lines is:
    1λ11λ2=2.7541×104cm1\dfrac{1}{\lambda_{1}} - \dfrac{1}{\lambda_{2}} = 2.7541 \times 10^{4} cm^{-1}
    Solution

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