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Atoms Test - 49

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Atoms Test - 49
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  • Question 1
    1 / -0
    Three photons coming from emission spectra of hydrogen sample are picked up. Their energies are $$12. 1eV, 10.2eV$$ and $$1.9eV$$ . These photons must come from
    Solution
     These energies are due the hydrogen electron dropping down from a higher excited energy level to a lower level .. so they must each be photons from a single atom (electrons don't move between atoms when they change energy levels) 

    12.1eV .. emitted when an electron drops from level n=3 (1.51 eV) to n=1 (13.60eV) 
    [excess energy 13.6 - 1.51 = 12.09eV] 

    10.2 eV .. emitted for a transition between n=2 (3.4eV) and n=1 (13.6eV) 

    1.9eV .. emitted for a transition between n=3 (1.51eV) and n=2 (3.40eV)
  • Question 2
    1 / -0
    A sodium atom emits a photon of wavelength $$590\ nm$$ and recoils with velocity $$v$$ equal to 
    Solution
    Here momentum is conserved.
    Momentum of photon rejected.
    $$\cfrac{h}{\lambda}=\cfrac{6.30\times 10^{-34}}{590\times 10^{-9}}$$
    Momentum of $$Na$$ atom
    $$=3.817\times 10^{-35}Kg\times Velocity$$
    $$\cfrac{h}{\lambda}=mv$$
    $$V=\cfrac{h}{\lambda m}=\cfrac{6.30\times 10^{-34}}{590\times 10^{-9}\times 3.817\times 10^{-35}}$$
    $$=0.029m/s$$
  • Question 3
    1 / -0
    The ratio of the speed of an electron in the first orbit of hydrogen atom to that in the first orbit of $$He^{+}$$ is
    Solution
    $$\begin{array}{l}\text { From single electron model we get - }\\\qquad\begin{aligned}\frac{m v^{2}}{r} &=\frac{k z e^{2}}{r^{2}}\\\Rightarrow \quad m v^{2} r =k z e^{2}\end{aligned} \\\text {Rearranging } \rightarrow \quad v \cdot(m v r)=k z e^{2}\end{array}$$


    $$\begin{array}{l}\text {  By Bohr quantization we have }\left[\text { mvr }=\frac{n h}{2 \pi}\right] \\\text { So putting in above eq we get, } \\\qquad \begin{array}{l}v \cdot \frac{n h}{2 \pi}=k z e^{2} \\\Rightarrow v=\frac{2 \pi k z e^{2}}{n h}\end{array}\end{array}$$

    $$\begin{array}{l}\Rightarrow \quad \frac{V_{H}}{V_{H e}}=\frac{Z_{H}}{n_{H}} \times \frac{n_{H e}}{Z_{H e}}=\frac{1}{1} \times \frac{1}{2}=\frac{1}{2} \\\Rightarrow \text { Velocity of hydrogen to velocity of Helium } \\\qquad V_{H}: V_{H e}=1: 2\end{array}$$


  • Question 4
    1 / -0
    In in nature there may not be an element for which the principal quantum number $$n> 4$$, then the total possible number of elements will be
    Solution
    It is based on electronic configuration
    $$1s^22s^22p^63s^23p^64s^24p^64d^{10}4f^{14}$$
    Since maximum number of electrons which can be arranged is 60, thus there are 60 elements can be exists in nature if principle quantum number >4 were not allowed.

  • Question 5
    1 / -0
    The electron in a hydrogen atom makes a transition $$n_1\rightarrow n_2$$ whose $$n_1$$ and $$n_2$$ are the principal quantum numbers of the two states. Assume the Bohr model to be valid. The frequency of orbital motion of the electron in the initial state is $${1/27}$$ of that in the final state. The possible value of $$n_1$$ and $$n_2$$ are

    Solution
    $$frequency \propto \cfrac{z^2}{n^3}$$
    $$\Rightarrow f=\cfrac{kz^2}{n^3}$$
    $$\therefore f_1=\cfrac{1}{27}f_2$$
    $$\Rightarrow \cfrac{kz_1^2}{n_1^3}=\cfrac{1}{27}\cfrac{kz_2^2}{n_2^3}\quad \left(z_1=z_2=1\right)$$
    $$\Rightarrow \cfrac{k}{n_1^3}=\cfrac{1}{27}\cfrac{k}{n_2^3}$$
    $$\Rightarrow n_1=3n_2$$
    $$\therefore n_1=3$$ and $$n_2=1$$
  • Question 6
    1 / -0
    In terms of Bohr radius $${a}_{0}$$, the radius of the second Bohr orbit of a hydrogen atom is given by:
    Solution
    Since, $$r \propto {n^2}$$
    Radius of $$2^{nd}$$ Bohr's orbit $$= 4a_0$$.
  • Question 7
    1 / -0
    An $$\alpha-$$ particle having energy $$10 MeV$$ collides with a nucleus of atomic number $$50$$.Then distance of closet approach will be
    Solution
    At the distance of closest approach the whole kinetic energy get converted into potential energy.
    So $$ KE=(1/4 \pi \epsilon_ 0)2Ze^2/r$$

    Where r is distance of closest approach.
    $$r=\dfrac{2Ze^2}{KE\times 4\pi \epsilon_0}$$
    putting $$Z=50, (1/4\pi \epsilon_0)=9\times10^9 ,e=1.6\times 10^{-19}, KE=10\times 1.6\times 10 ^{-13} Joule$$
    we get $$r=14.4\times 10^{-15}m$$
    Option D is correct.
  • Question 8
    1 / -0

    The key feature Of Bohr's theory Of spectrum Of hydrogen atom is the quantization Of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy Of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition.A diatomic molecule has moment Of inertia I. By Bohr's quantization condition its rotational energy in the nth level (n=0 is not allowed) is 

    Solution
    $$\begin{array}{l}\text { Given - * A Diatomic molecule- } \\\text { let, the structure of the molecule be like- }\end{array}$$

    $$\Rightarrow I=2 m r^{2}$$

    $$\begin{array}{l}\text { we howe Rotational encrgy- } \\\qquad E=\frac{1}{2} I \omega^{2}=m \omega^{2} r^{2}-(1) \\\text { By Bohr quantization of angular momentum } \\\qquad 2 \text { mwr}^2=\frac{n h}{2 \pi} \quad-(2) \\\text { From } e q(1) \text { and (2) we get ,}\\\qquad E=\frac{n^{2} h^{2}}{16 \pi^{2} m r^{2}}=\frac{n^{2} h^{2}}{8\pi^{2} I}\end{array}$$

  • Question 9
    1 / -0
    In Bohr's model of hydeogen atom, let PE represent potential energy, and TE the total energy. In going to a higher level:
    Solution
    In Bohr's model of hydrogen atom PE is given by,
     
    $$\dfrac { -kz{ e }^{ 2 } }{ r } $$

    Here we go from lower level to higher level, which implies that radius increases, which implies that numerically PE of system decreases.
    But due to negative sign it actually increases.Hence when going to higher energy level, PE increases.

    And as we know TE is given by,
     
    $$\dfrac { -2{ \pi  }^{ 2 }m{ k }^{ 2 }{ z }^{ 2 }{ e }^{ 4 }\qquad  }{ { n }^{ 2 }{ h }^{ 2 } } $$

    And as we go into higher level, n increases and TE numerically decreases. But due to negative sign , it increases.

    Ans: PE increases, TE increases.
  • Question 10
    1 / -0
    An electron in hydrogen atom makes a transition $$n_{1}\rightarrow n_{2}$$ where $$n_{1}$$ and $$n_{2}$$ are principal quantum numbers of the two states. Assuming Bohr's model to be valid, the time period of the electron in the initial state is eight times that in the final state. The possible values of $$n_{1}$$ and $$n_{2}$$ are
    Solution
    $$\begin{array}{l}\text { Given- * An electron makes transition } n_{1}\longrightarrow n_{2} \\\text { such that Time period in initial state is } \\8 \text { times the final state. }\left[T_{1}=8T_{2}\right]\end{array}$$
    $$\text{We know that in a single $e^{-}$ model-}$$
    $$T \propto \frac{n^{3}}{z^{2}}$$

    $$\Rightarrow \frac{T_{1}}{T_{2}}=\frac{n_{1}^{3}}{n_{2}^{3}}=\frac{8}{1}$$

    $$\begin{aligned}\Rightarrow \frac{n_{1}}{n_{2}}=2 \Rightarrow n_{1}=4\quad n_{2}=2 \\& \text { is one of a possible value. }\end{aligned}$$
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