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Atoms Test - 48

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Atoms Test - 48
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  • Question 1
    1 / -0
    A single electron orbits around a stationary nucleus of charge $$+Ze$$, where $$Z$$ is a constant and $$e$$ is the magnitude of the electronic charge. It required $$47.2\ eV$$ to excite the electron from the second Bohr orbit to the third Bohr orbit. Find the value of $$Z$$.
    Solution

  • Question 2
    1 / -0
    The total energy of in the electron in the hydrogen atom in the ground state is $$-13.6\ eV$$. Which of the following is its kinetic energy in the first exited state?
    Solution
    Energy nth orbit of hydrogen atom is given by,

    $$E_n=\dfrac{-13.6}{n^2}e.V$$

    For ground state, $$n=1$$

    $$\therefore E_l=\dfrac{-13.6}{l^2}=-13.6eV$$

    for first excited state, $$n=2$$

    $$\therefore E_2=\dfrac{-136.6}{2^2}=\dfrac{-136.6}{4}=-3.4eV$$

    Kinetic energy of an electron in the first excited state is,

    $$K.E=-E_2$$

    $$\Rightarrow K.E=3.4eV$$
  • Question 3
    1 / -0
    If the attractive potential between electron and proton were given by $$V(r) = V_{0}ln \left (\dfrac {r}{r_{0}}\right )$$ (where $$V_{0}, r_{0}$$ are constants), the radius $$r_{n}$$ of the $$nth$$ Bohr orbit in Hydrogen atom would be proportional to
    Solution
    $$V_{(r)} = V_0 \ln \left(\dfrac{r}{r_0} \right)_{nth}$$
    $$r_n \alpha ?$$
    $$r_n =$$ radius of both orbit in $$H-$$ atom
    $$F = -\dfrac{dv}{dr} = \dfrac{mv^2}{r}$$
    $$v = \sqrt{\dfrac{Fr}{m}}$$
    and $$mv_n \ln = \dfrac{nh}{2 \pi} $$ (angular momentum)
    $$\ln \, 2n$$
  • Question 4
    1 / -0
    If element with principal quantum number $$n > 4$$ were not allowed in nature, the number of possible elements would be
    Solution
    If the element with principal quantum
    number$$n>4$$
    when $$n=4$$
    The formula becomes$$=2(4)^2\\=16\times2\\=32$$
    When $$n=5$$
    The formula becomes$$=2(5)^2$$
    Hence the number of elements would$$=60$$

  • Question 5
    1 / -0
    A single electron orbits around a stationary nucleus of charge $$+Ze$$, where $$Z$$ is a constant and $$e$$ is the magnitude of the electronic charge. It required $$47.2\ eV$$ to excite the electron from the second Bohr orbit to the third Bohr orbit.
    Find The wavelength of the electromagnetic radiation required to remove the electron from the first Bohr orbit.
    Solution

  • Question 6
    1 / -0
    The number of waves in the third orbit of H atom is
    Solution
    No of waves= Shell No of the orbit= 3 
  • Question 7
    1 / -0
    The radius of the first Bohr orbit is $$a_{0}$$. The $$n^{th}$$ orbit has a radius:
    Solution
    Radius of $$n^{th}$$ orbital
    $$r_n=\dfrac{\epsilon_0n^{2}h^{2}}{\pi mZe^2}$$
    Wherein 
    $$=r_n\propto\dfrac{n^2}{Z}$$

    $$=\dfrac{\epsilon_0h^{2}}{\pi me^2}=0.529$$

    $$r=\dfrac{n^2}{Z}a_0$$

    For $$Z=1$$ $$r=n^{2}a_0$$
  • Question 8
    1 / -0
    The average current due to an electron orbiting the proton in the $$n^{th}$$ Bohr orbit of the hydrogen atom is
    Solution
    We know,
    $$r_n \propto \dfrac{1}{n^2}$$

    $$v_n \propto \dfrac{Z}{n}$$

    $$\omega=\dfrac{2\pi r_n}{v_n}$$

    $$\omega\propto\dfrac{Z}{n^3}$$

    $$I\propto\omega\propto\dfrac{1}{n^{-3}}$$

    $$L\propto n^{-3}$$

    Hence option $$\textbf C $$  is correct.
  • Question 9
    1 / -0
    The ionisation potential of a hydrogen atom is $$13.6\ volt$$. The energy required to remove an electron from the second orbit of hydrogen is:
    Solution
    Energy required to remove electron in the $$n=2$$ state 

    $$E_n=-13.6\dfrac{Z^2}{n^2}$$

    $$\Delta E=-13.6\left[\dfrac{1}{\infty^2}-\dfrac{1}{4}\right]$$

    $$\Delta E=\dfrac{13.6}{4}eV$$

    $$=3.4\ eV$$
  • Question 10
    1 / -0
    The radius of the shortest orbit in a one-electron system is $$18\ pm$$. It may be
    Solution

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