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Atoms Test - 12

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Atoms Test - 12
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  • Question 1
    1 / -0
    What would be the radius of second orbit of He+{He}^{+} ion?
    Solution
    Radius of the nth orbit, rn=0.529 n2Zr_n = 0.529  \dfrac{n^2}{Z}   AoA^o
    For He+He^+ ion, Z=2Z = 2
    Radius of 2nd orbit, r2=0.529222r_2 = 0.529\dfrac{2^2}{2}
         r2 = 1.058\implies   r_2  =  1.058 A˚\mathring A
  • Question 2
    1 / -0
    1. According to Bohr's theory, the radius of the nthn^{th} orbit of an atom of atomic number Z Z is proportional to 
    Solution
    According to Bohr's theory, the coulomb force on the electron causes its centripetal acceleration.
        14πϵ0Ze2r2=mv2r\implies \dfrac{1}{4\pi\epsilon_0}\dfrac{Ze^2}{r^2}=\dfrac{mv^2}{r}
    Also one of Bohr's postulates states mvr=nh2πmvr=\dfrac{nh}{2\pi}
    Thus eliminating vv from the equations gives
        rn2Z\implies r\propto \dfrac{n^2}{Z}
  • Question 3
    1 / -0
    Light from a tungsten filament lamp gives
    Solution
    Light from a tungsten filament lamp has an emission spectrum which is continuous in nature.
  • Question 4
    1 / -0
    Check the wrong statement :
    Solution
    Line respectrum is characteristic of the element because it obtained of atoms only. Continous spectrum is characteristics of the source of light.
    In the spectrum of sodium, there are two prominent yellow lines of wavelength 589.0 nm and 589.6 nm.
    Absorption line spectrum is not characteristic of the element.
  • Question 5
    1 / -0
    The element which was observed in solar spectrum is 
    Solution
    Pierre Janssen discovered the gaseous nature of the solar chromosphere and with some justification the element helium.
  • Question 6
    1 / -0
    Match the appropriate pairs from Lists I and II:
    List-IList - II
    a) Nitrogen moleculee) Continuous spectrum
    b) Incandescent solidsf) Absorption spectrum
    c) Fraunhofer linesg) Band spectrum
    d) Electric arc between iron rodsh) Emission spectrum
    Solution
    Nitrogen molecules shows band spectrum
    Incandescent solids shows continuous spectrum Fraunhofer lines are missing lines in the absorption spectrum
    Eletric are between iron rods produces llight beam which is when dispersed we get emission spectrum.
  • Question 7
    1 / -0
    The spectra used to identify the elements in the mixture is :
    Solution
    To identify the element in the mixture we used emission and obsorption spectra.
  • Question 8
    1 / -0
    Solar spectrum is an example of 
    Solution
    Solar spectrum is a line absorption spectrum which is also called as Fraunhofer lines of missing wavelengths.
  • Question 9
    1 / -0
    The idea of quantum nature of light has emerged in an attempt to explain :
    Solution
    Black body radiation could not be explained by the classical theory.
    The Quantum nature of light has emerged in an attempt to explain black body radiation with the insertion of Planck's constant and thus the Planck's law.
    I(V,T)=2hv3c21ehvkT1I(V,T)=\dfrac{2hv^{3}}{c^{2}}\dfrac{1}{e^{\frac{hv}{kT}}-1}
  • Question 10
    1 / -0
    Consider the following two statements A and B and identify the correct choice in the given answers
    A) Line spectra is due to atoms in gaseous state
    B) Band spectra is due to molecules 
    Solution
    The light emitted by one kind of atoms generally have widely separated wavelength components. When such a light is dispersed, we get certain sharp bright lines on a dark background. Such a spectrum is called line emission spectrum.The wavelength emitted by the molecules are grouped, each group being well separated from the other. The wavelengths in a group are close to one another and appear as continuous.The spectrum looks like separate bands of varying colors. Such a spectrum is called a band emission spectrum.So, line spectra are due to atoms in a gaseous state and band spectra are due to molecules.
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