Self Studies

Atoms Test - 32

Result Self Studies

Atoms Test - 32
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    If the radius of first Bohr orbit is rr, then the radius of second orbit will be
    Solution
    The radius of an orbit of revolution of electron is:
    r=n2h2ϵ0πme2r = \dfrac{n^2h^2\epsilon_0}{\pi m e^2}
    where, variables have their usual meanings.
    rn2\Rightarrow r \propto n^2

    Therefore, r1n12r_1 \propto n_1^2 and r2n22r_2 \propto n_2^2
    r2r1=n22n12\Rightarrow \dfrac{r_2}{r_1} = \dfrac{n_2^2}{n_1^2}

    r2r=222112\therefore \dfrac{r_2}{r} = \dfrac{2_2^2}{1_1^2}

    r2=4r\therefore r_2 = 4r
  • Question 2
    1 / -0
    Whenever a hydrogen atom emits a photon in the Balmer series
    Solution
    Balmer series, electron is still in excited state. (n=2n=2), so it will definitely come to a lower energy state by transit from (n=2ton=1n=2 to n=1)

  • Question 3
    1 / -0
    Two electrons in an atom are moving in orbit of radii RR and 9R9R respectively. The ratio of their frequencies will be
    Solution
    Radius is directly proportional to square of nn. Therefore, ratio of principal quantum number is 1:31:3
    Hence, ratio of frequencies is 27:127:1.
  • Question 4
    1 / -0
    The electron in a hydrogen atom jumps from the ground state to the higher energy where its velocity is reduced to one-third of its initial value. If the radius of the orbit in the ground state is rr, the radius of new orbit will be
    Solution
    Velocity is inversely proportional to nn, hence n2=3{ n }_{ 2 }=3
    Radius is directly proportional to square of nn.
    Therefore,  r3=9r r_{ 3 }=9r
  • Question 5
    1 / -0
    The radius of the Bohr orbit in the ground state of hydrogen atom is 0.5A˚0.5\mathring{A}. The radius of the orbit of the electron in the third excited state of He+He^+ will be
    Solution
    Radius, rn2Zr\propto \dfrac{n^2}{Z}

    Thus, r2r1=n22Z1n12Z2=42×112×2=8\dfrac{r_2}{r_1}=\dfrac{n_2^2Z_1}{n^2_1Z_2}=\dfrac{4^2\times 1}{1^2\times 2}=8

    r2=8r1=8(0.5)=4 A˚\Rightarrow r_2=8r_1=8(0.5)=4  \mathring A
  • Question 6
    1 / -0
    In the above figure, E1E_1 to E6E_6 represent some of the energy levels of an electron in the hydrogen atom.
    Which one of the following transitions produces a photon of wavelength in the ultra-violet region of the electromagnetic spectrum?

    Solution
    If a transition is occurred into E1{ E }_{ 1 }, resulting radiation will be U.V
  • Question 7
    1 / -0
    The approximate value of quantum number n 'n\ ' for the circular orbit of hydrogen of 0.0001 mm0.0001\space mm in diameter is
    Solution
    For d= 0.0001 mm=107m =1000 Ad =  0.0001\ mm = 10^{-7} m  = 1000\ A
    r=0.529n2Z r = 0.529 \cfrac{n^2}{Z}
    For hydrogen, Z=1Z =1
    500=0.529n2 500 = 0.529 n^2
    n=30.74n = 30.74
  • Question 8
    1 / -0
    How many revolutions does an electron complete in one second in the first orbit of hydrogen atom?
    Solution
    The radius of first orbit , r1=0.53×1010mr_1=0.53 \times 10^{-10} m

    Thus, circumference of this orbit , p=2πr1=2π×(0.53×1010)=3.33×1010mp=2\pi r_1=2\pi\times (0.53\times 10^{-10})=3.33\times 10^{-10} m

    The speed of first orbit is approximately , v1=c/137=3×108137=2.19×106m/sv_1=c/137=\dfrac{3\times 10^8}{137}=2.19\times 10^6 m/s

    Time period, T=p/v1=3.33×10102.19×106=1.52×1016sT=p/v_1=\dfrac{3.33\times 10^{-10}}{2.19\times 10^6}=1.52\times 10^{-16} s 

    The number of revolutions per sec is the frequency. 
    So, the frequency of the revolutions, f=1T=11.52×1016=6.6×1015f=\dfrac{1}{T}=\dfrac{1}{1.52\times 10^{-16}}=6.6\times 10^15 
  • Question 9
    1 / -0
    Which of the following parameters are the same for all hydrogen-like atoms and ions in their ground states?
    Solution
    According to Bohr's model of an atom, energy of  nth orbit is given by: En=13.6 Z2n2  eVE_n = -13.6   \dfrac{Z^2}{n^2}    eV
    Radius of nth orbit, rn=0.529 n2Z   Aor_n = 0.529   \dfrac{n^2}{Z}       A^o
    Speed of an electron in nth orbit, vn = 2.18 ×106 Zn  m/sv_n  =  2.18  \times 10^6   \dfrac{Z}{n}    m/s
    And  orbital angular momentum, Ln =nh2πL_n  =\dfrac{nh}{2\pi}
        \implies Radius of the orbit, speed of the electron, and energy of an atom depends on ZZ whereas orbital angular momentum is independent of ZZ.
    Hence, orbital angular momentum remains the same for all hydrogen like atoms and ions in ground state  (n=1)(n=1).
  • Question 10
    1 / -0
    The electric potential energy between a proton and an electron is given by U=U0lnrr0U=U_0ln\dfrac{r}{r_0}, where r0r_0 is a constant. Assuming Bohr's model to be applicable, write the variation of r 'r\ ' with nn, nn being the principal quantum number.
    Solution
    The electrostatic interaction force between electron and proton is F=dUdr=U0rr0F=\dfrac{dU}{dr}=\dfrac{U_0}{rr_0}
    According Bohr's first postulate, mv2r=F=U0rr0\dfrac{mv^2}{r}=F=\dfrac{U_0}{rr_0}
    or v2=U0mr0constantv^2=\dfrac{U_0}{mr_0} \rightarrow constant
    According Bohr's second postulate, mvrn=nh2πmvr_n=\dfrac{nh}{2\pi}
    As m,v and h are constant so we can write, rnnr_n\propto n
Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now