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Atoms Test - 74

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Atoms Test - 74
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  • Question 1
    1 / -0
    An electron collides with a fixed hydrogen atom in its ground state. Hydrogen atom gets excited and the colliding electron loses all its kinetic energy. Consequently the hydrogen atom may emit a photon corresponding to the largest wavelength of the Balmer series. The main KE of colliding electron will be
    Solution

  • Question 2
    1 / -0
    An alpha particle collides elastically with a stationary nucleus and continues moving at an angle of $$60^0$$ with respect to the original direction of motion. The nucleus recoils at an angle of $$30^0$$ with respect to this direction. Mass number of the nucleus is  :
    Solution
    Given collision of $$\alpha$$-particle with nucleus is elastic.
    Mass of $$\alpha$$ particle$$=4$$ units
    mass of neucleus$$=$$m
    Given  $$\theta =60^o$$
        $$\phi =30^o$$
    $$\phi +\theta =90^o$$
    as we know that when a particle moving with velocity v colloid obliquely & elastically with a particle of same mass and at rest then after collision angle between their velocities is $$\dfrac{\pi}{2}$$
    $$\Rightarrow$$ Mass of $$\alpha$$ particle$$=$$mass of nucleus
    $$m=4$$ unit.
  • Question 3
    1 / -0
    Choose the correct statement (s) for hydrogen and deuterium atoms considering motion of reaction
    Solution
    $$ \begin{array}{l} \text { Bohr's considerd the nucleus is at rest and only electron is revolving around it. } \\ \text { but here in question motion of nucleus is given.} \end{array} $$ 
    $$ \begin{array}{l} \text { Hydrogen }= _{1}{H}^{2} \\ \text { deuterium }= _1{\mathrm{H}^{2}} \end{array} $$
     $$ \begin{array}{l} \text { Now, if we consider motion of nucleus } \\ \text { also, then we have to consider nucleus mass } \\ \text { also. } \end{array} $$ 
    $$ \text{  (a)} \text { Now, } r  \propto \dfrac{n^{2}}{m} $$
     $$ \begin{array}{l} \text { So, radius of deuterium is less than that of hydrogen, option (a) is correct } \end{array} $$ 
    $$ \text { (b) velocity } \propto \dfrac{z^{2}}{n^{2} m} $$ 
    $$ \begin{array}{l} \text { So, velocity of electron in hydrogen is more, option B incorrect } \end{array} $$
     
    $$ \begin{array}{l} \text { (c) wavelength }\\ \dfrac{1}{\lambda}=R z^{2}\left(\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}\right) \\ \text { and } \therefore R \propto m \end{array} $$ 
    $$ \begin{aligned} & \frac{1}{\lambda} \propto R \\ &=\frac{1}{\lambda} \propto m \end{aligned} $$ 
    $$ \begin{array}{l} \text { So, wavelength of hydrogen will be more than that of deuterium, } \end{array} $$ 
    $$ \text { option c is incorrect } $$
     $$ \begin{array}{l} \text { (d) as angular momentum is conserved always inspite of mass and atomic no. , it only depend on Energy level. } \\ \text { mvr }=\dfrac{n h}{2 \pi} \end{array} $$
     $$ \text { So, option }(d) \text { is incorrect. } $$.
  • Question 4
    1 / -0
    An electron jumps from the $$4^{th}$$ orbit to the $$2^{nd}$$ orbit of hydrogen atom $$(R=10^{5}\ Cm^{-1})$$. Frequency is in $$Hz$$ of emitted radiation will be
    Solution
    $$\begin{array}{l}\text { Electron jumps from } 4^{\text {th }} \text { orbit to } \\2^{\text {nd }} \text { orbit. } \\\text { Let wavelength be "x" then } \\\qquad \frac{1}{\lambda}=R Z^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\end{array}$$

    $$\text { Here, } \begin{aligned}& R=10^{5} \mathrm{~cm}^{-1} \\& z=1\\& n_{1}=2, n_{2}=4 \\\frac{1}{\lambda} &=10^{5}\left[\frac{1}{2^{2}}\frac{1}{4^{2}}\right] \\\frac{1}{\lambda}&=10^{5}\left[\frac{1}{4}-\frac{1}{16}\right] \\\frac{1}{\lambda} &=10^{5}\left[\frac{3}{16}\right] \\\frac{1}{\lambda} &=\frac{3 \times 10^{5}}{16}-(1)\end{aligned}$$


    $$\begin{array}{l}\text { Frequency is } \frac{c}{\lambda} \\c=3 \times 10^{8} \mathrm{~m}=3 \times 10^{10} \mathrm{~cm} \\\text { From frequency, } v=3 \times 10^{10} \times \dfrac{3 \times 10^{5}}{16}\\\qquad v=\dfrac{9 \times 10^{15}}{16}\end{array}$$

  • Question 5
    1 / -0
    According to Bohr's theory of the hydrogen atom, the speed $$v_n$$ of the electron in a stationary orbit is related to the principal quantum number n as (C is a constant).
    Solution

  • Question 6
    1 / -0
    In a Geiger-Marsden experiment if the distance of closest approach is $$d$$ to the nucleus and an $$\alpha$$ particle of energy E then
    Solution

  • Question 7
    1 / -0
    The energy levels of a certain atom for 1st, 2nd and 3rd levels are E, 4E$$_{1\to3}$$ and 2E respectively. A photon of wavelength $$\lambda $$ is emitted for a transition $$3\rightarrow 1$$. What will be the wavelength of emissions for transition $$2\rightarrow 1$$:
    Solution
    Let $$E_{1}, E_{2}, E_{3}$$ represents the energy of $$1st, 2nd$$ and $$3rd$$ energy levels respectively such that 
    $$E_{1}=E$$
    $$E_{2}= 4E$$
    $$E_{2}= 2E$$
    Now, we know that during transition of electron from higher energy level to lower energy level, energy in released in the form of photon having energy equal to the energy difference between two energy levels.
    Ob transition from $$3 \rightarrow 1$$ energy level. 
    Energy of photon released $$= E_{3}- E_{1} $$--------$$(1)$$
    As we know that energy of photon is given by $$=\dfrac{hc}{\lambda}$$
    where $$\lambda $$ wavelength of photon 
    from $$(1)$$
    $$\dfrac{hc}{\lambda} = E_{3}-E_{1}=E$$
    $$\Rightarrow \lambda= \dfrac{hc}{E}$$-------$$(2)$$
    On transition from $$2 \rightarrow 1$$ energy level.
    Energy of photon released $$=E_{2}-E_{1}$$--------$$(2)$$
    Let, wavelength of photon released be $$\lambda$$
    $$\Rightarrow $$ Energy of photon released $$=\dfrac{hc}{\lambda}$$
    From $$(3)$$
    $$E_{2}-E_{1}= \dfrac{hc}{\lambda ,}$$
    $$\Rightarrow 3E= \dfrac{hc}{\lambda ,}$$
    $$\Rightarrow \lambda ' = \dfrac{hc}{3E} =\dfrac{1}{3} \left( \dfrac{hc}{E} \right)$$
    $$\Rightarrow \lambda ' = \dfrac{1}{3} (\lambda) (from \ (2))$$
    $$\Rightarrow \lambda ' = \dfrac{\lambda}{3}$$
    Hence, wavelength of photon emitted for transition $$2 \rightarrow 1$$ in $$\dfrac{\lambda}{3}$$.
  • Question 8
    1 / -0
    Assuming Bohr's model for $$Li^ {++}$$ atom, the first excitation energy of ground state of $$Li^ {++}$$ atom is
    Solution
    $$ \begin{array}{l} \text { Given, } \\ Li^{++} \text {  atom: } \\ \text { first excitation Energy of ground state } L i^{++} \\ \text {mean excitation of electrom from } \\ \qquad n=1 \text { to } n=2 \end{array} $$ 
    $$ \begin{array}{l} \text { we know, } \\ \text { Energy }=13.6 z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \end{array} $$
     $$ \begin{aligned} &=13.6 \times 9 \times\left(\frac{1}{1}-\frac{1}{4}\right) \\ E &=\frac{13.6 . \times 9 \times 3}{4} \mathrm{ev} \\ E &=91.8 \mathrm{ev} \end{aligned} $$
  • Question 9
    1 / -0
    In a hypothetical system a particle of mass m and charge -3 q is moving around a very heavy particle having charge q. Assuming Bohr's model to be true to this system, the orbital velocity of mass m when it is nearest to heavy particle is:
    Solution

  • Question 10
    1 / -0
    A hydrogen atoms emits green light when it changes from the n=4 energy level to n=2 level. which color of light would the atom emit when it changes from the n=5 level to n=2 level ?
    Solution
    According to Balmer series.
    $$\overline { V } =R\left[ \dfrac { 1 }{ { z }^{ 2 } } -\dfrac { 1 }{ { n }^{ 2 } }  \right] $$
    $$n=3,4,5,6,......$$
    Spectra basically is belonging the location of visible zone. When wavelength is in between $$4000{ A }^{ 0 }$$ to $$5000{ A }^{ 0 }$$ then, green color of light would the atom emit. When it changes from the $$n=5$$ level to $$n=2$$ level.
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