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Atoms Test - 75

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Atoms Test - 75
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  • Question 1
    1 / -0
    Let the potential energy of a hydrogen atom in the ground state be zero. Then, its energy in the first excited state will be  
    Solution

  • Question 2
    1 / -0
    If the ionisation energy of a hydrogen like Bohr atom is 4 Rydberg, then the wavelengths of radiation emitted when the electron jumps from first excited state to the ground state and the radius of the first orbit of this atom are:
    Solution
    $$\begin{array}{l}\text { Ionisation energy, } E=R z^{2}\\\qquad\begin{aligned}Rz^{2} &=4 R \\z &=2\end{aligned}\end{array}$$


    $$\text { Radius of 1st orbit is } \dfrac{0.529}{2}A^{0}\text{=0.265} A^{0}$$

    $$\begin{array}{l}\text { wavelength } \\\qquad \dfrac{1}{\lambda}=R z^{2}\left[\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}\right]\end{array}$$


    $$\begin{array}{ll}n_{1}=1 & n_{2}=2 \\\text { we get, } & \lambda=304 A^{0}\end{array}$$
  • Question 3
    1 / -0
    The ratio of areas within the electron orbits for the first excited state to the ground state for hydrogen atom is
    Solution
    The square radius of the orbit is directly proportional to the power of four of the  number of state.
    $$r^2 \alpha$$  $$n^4$$
    Where, n is the number of state
    Areas of hydrogen for excited state
    $$A_1$$ $$\alpha$$ $$n_1^4$$
    Here, n = 2 for excited state
    The area of hydrogen for the ground state
    $$A_0$$ $$\alpha$$ $$n_0^4$$
    Here, n = 1 for ground state
    Now, the ratio of areas between the electron orbits for the first excited state to the ground state for the hydrogen atom is
    $$\frac{A_1}{A_0} = (\frac{n_1}{n_0})^4$$
    $$\frac{A_1}{A_0} = (\frac{2}{1})^4$$
    $$\frac{A_1}{A_0} = (\frac{16}{1})$$
    Hence, The ratio of the area is $$16:1$$.
  • Question 4
    1 / -0
    The innermost orbit of the hydrogen atom has a diameter of $$1.06 A^o$$. What is the diameter of the tenth orbit?
    Solution

  • Question 5
    1 / -0
    Find an expression for de Broglie wavelength for an electron in the $$n^{th}$$ orbit of the Bohr model of the hydrogen atom and hence find its value when the electron is in n = 4 level.
    Take radius of nth orbital of Bohr's model of H-atom; $$r_n = (53 \ n^2) $$ pm
    Solution

  • Question 6
    1 / -0
    A satellite of mass $$10\ kg$$ orbites around the earth once every $$2\ h$$ in an orbit of radius $$8000\ km$$. If Bohr's angular momentum is assumed to be applicable to the satellite, the quantum number of the orbit of the satellite is
    Solution

  • Question 7
    1 / -0
    According to Bohr modal, magnetic field at the centre (at the nucleus) of a hydrogen atom due to the motion of electron in $$n^{th}$$ orbit is proportional to:
    Solution

  • Question 8
    1 / -0
    The $$H_{a}$$-line of hydrogen
    Solution

  • Question 9
    1 / -0
    The wave number of first line in Balmer series of hydrogen spectrum is (Rydberg constant, $$R_H = 109,687 \,cm^{-1}$$) nearly
    Solution
    $$\begin{array}{l}\text { Wave number } \\\text { Here, }\quad\left[\bar{v}=R_{H} z^{2}\left[\frac{1}{n_{i}{ }^{2}}-\frac{1}{n_{2}^{2}}\right]\right. \\R_{H}=109687 \\z=1\end{array}$$

    $$ \begin{aligned}\text { Balmar } n_{1}=2 \text { , and } n_{2}=3\\\bar{v}=109687\left[\frac{1}{(2)^{2}}-\frac{1}{(3)^{2}}\right]\\\bar{v}=109687\left[\frac{1}{4}-\frac{1}{9}\right]\end{aligned}$$

    $$\bar{v}=15200 \mathrm{~cm}^{-1}$$
  • Question 10
    1 / -0
    Which of the following is inversely proportional to the square of principal quantum number in Bohr's atomic model?
    Solution

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