Self Studies

Atoms Test - 79

Result Self Studies

Atoms Test - 79
  • 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
    The ratio of the largest to shortest wavelength in Lyman series of hydrogen spectra is
    Solution
    For Lyman series $$\dfrac {1}{\lambda_{max}}=R \left[ \dfrac {1}{1^2}-\dfrac {1}{2^2}\right]=\dfrac 34\ R$$ and 
    $$\dfrac {1}{\lambda_{min}}=R\left[ \dfrac {1}{1^2}-\dfrac{1}{\infty^2}\right]=\dfrac R1$$
    $$ \Rightarrow \dfrac{\lambda_{max}}{\lambda_{min}}=\dfrac {4}{3}$$
  • Question 2
    1 / -0
    The wavelength of the least energetic photon in the Balmer spectrum of hydrogen atom is:
    Solution

  • Question 3
    1 / -0
    The hydrogen atom in ground state is excited by a monochromatic radiation of $$  \lambda=975 A^{\circ}  $$ . Number of spectral lines in the resulting spectrum emitted will be
    Solution

  • Question 4
    1 / -0
    An excited electron of H-atoms emits a photon of wavelength λ and returns in the ground state, the principal quantum number of excited state is given by:
  • Question 5
    1 / -0
    The radius of which of the following orbit is same as that of the first Bohr's orbit of hydrogen atom 
  • Question 6
    1 / -0
    The relation between $${\lambda}_{1}$$: wavelength of series limit of Lyman series, $${\lambda}_{0}$$: the wavelength of the series limit of Balmer series and $${\lambda}_{3}$$: the wavelength of first line of Lyman series is:
    Solution

  • Question 7
    1 / -0
    In hydrogen atom electron is revolving around nucleus in circular orbit with frequency of $$6.57 \times {1^{15}}$$ revolutions. If its equivalent magnetic moment is $$9.24 \times {10^{ - 24}}A{m^2}$$ , the radius of orbit of electron is 

    Solution

  • Question 8
    1 / -0
    According to the Bohr theory of the hydrogen atom electrons starting in the $$4^{th}$$ energy level and eventually ending in the ground state could produce a total of how many lines in the hydrogen spectra ?
  • Question 9
    1 / -0
    Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line H-atom is suitable for this purpose ?
    Given: $${ R }_{ H }=1\times { 10 }^{ 5 }cm, h=6.610^{ -34 }Js, C=3\times 10^{ 4}m/{ s }^{ -1 }$$ 
    Solution

  • Question 10
    1 / -0
    The frequency of first line of Balmer series in hydrogen atom is $$v_0$$. The frequency of corresponding line emitted by singly ionised helium atom is -
    Solution
    $$v \propto Z^2 \Rightarrow \dfrac{v_{H_2}}{v_{He}}=\left( \dfrac 12 \right)^2=\dfrac 14 $$
    $$\Rightarrow v_{He}=4v_{H_2}=4v_0$$
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