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Atoms Test - 56

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Atoms Test - 56
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  • Question 1
    1 / -0
    When hydrogen atom is in its first excited level, its radius is how many times its ground state radius:
    Solution
    $$r \propto n^2$$. For ground state $$n=1$$ and for first excited state $$n=2$$
  • Question 2
    1 / -0
    When hydrogen atom is in its first excited level, its radius is _______ of the Bohr radius.
    Solution
    When hydrogen atom is in its first excited level,
    $$n=2$$ 
    The radius of hydrogen atom in the first excited level is
    $$r=n^2r_0$$
    $$r=(2)^2r_0=4r_0$$ 
    where, $$r_0=$$ Bohr's radius.
    The correct option is B.
  • Question 3
    1 / -0
    In a beryllium atom, if $$a_{0}$$ be the radius of the first orbit, then the radius of the second will be in general.
    Solution
    $$F\propto n^2$$

    $$r=\dfrac{n^2h^2\epsilon _0}{\pi mZe^2}$$

    $$r_1=\dfrac{h^2\epsilon _0 }{\pi mZe^2}=a_0$$

    $$r_n=\dfrac{n^2h^2\epsilon_0}{\pi mZe^2}$$

    $$r_n=a_0n^2$$

  • Question 4
    1 / -0
    The colour which corresponds to $$\mathrm { H } _ { \mathrm { B } } ( 4 \rightarrow 2 )$$ linein the Balmer series of hydrogen spectra is:
  • Question 5
    1 / -0
    A proton and an $$\alpha$$ particle having equal kinetic energy are projected in a uniform transverse electric field as shown in figure

    Solution
    Equal kinetic energy.
    But the trajectory will depend on $$acceleration $$ which is $$F=\dfrac{qE}{m}$$ Which is $$\dfrac{eE}{m_p}$$ for proton and  $$\dfrac{2eE}{4m_p}=\dfrac{eE}{2m_p}$$ for $$\alpha $$ particle.
     $$E$$ is electric field , $$q$$ is charge and $$m_p$$ is mass of $$one $$ proton.
    $$\text{As they are opposite in charge so opposite direction and proton having more acceleration will show}$$
    $$\text{ more curved trajectory.  }$$
    Option A is correct.
  • Question 6
    1 / -0
    In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is $$\lambda$$. If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be?
    Solution
    For M $$\rightarrow$$ L steel
    $$\dfrac{1}{\lambda}=K\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)=\dfrac{K\times 5}{36}$$
    for N$$\rightarrow$$L
    $$\dfrac{1}{\lambda'}=K\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)=\dfrac{K\times 3}{16}$$
    $$\lambda'=\dfrac{20}{27}\lambda$$.
  • Question 7
    1 / -0
    A particle moves along X-axis as $$x=4\left( t-2 \right) +a{ \left( t-2a \right)  }^{ 2 }$$ Which of the following is true?
    Solution

  • Question 8
    1 / -0
    the ground state energy of hydrogen  atom is -13.6 eV . the energy  of second excited state of $$He^{ + }$$ ion in eV is : 
    Solution
    $$\begin{array}{l} E=-13.6\times \dfrac { { { z^{ 2 } } } }{ { { n^{ 2 } } } }  \\ =-13.6\times \dfrac { 4 }{ 9 }  \\ =-6.04\, eV \end{array}$$
    Hence,
    option $$(D)$$ is correct answer.
  • Question 9
    1 / -0
    Imagine an atom made up of proton and a a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength $$\lambda $$ (give in term of the Rydberg constant R for the hydrogen atom) equal to
    Solution
    Energy is related to mass
    $${R_n} \propto m$$
    The longest wavelength λmaxλmax photon will correspond to the transition of particle from$$n=3$$ to$$ n=2$$
    $$\begin{array}{l} \dfrac { 1 }{ { { \lambda _{ \max   } } } } =2R\left( { \dfrac { 1 }{ { { 2^{ 2 } } } } -\dfrac { 1 }{ { { 3^{ 2 } } } }  } \right)  \\ { \lambda _{ \max   } }=\dfrac { { 18 } }{ 5 } R \end{array}$$
    Hence, Option $$C$$ is correct.
  • Question 10
    1 / -0
    In Bohr series of lines of hydrogen spectrum, third line from the red end corresponds to which one of the following inner orbit jumps of electron for Bohr's orbit in atom of hydrogen 
    Solution
    We see red end means it is a part of the visible region and obviously only Balmer series corresponds to the visible region for the Balmer series $$n_1=2$$ and red end means low energy or third end from this means $$n_2=-5 $$
    $$2(r)5. 3$$
    $$2(r)4. 2 $$
    $$2(r)3. 1$$ 
    The correct answer is (D) $$5 \to 2$$
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