a. What quantum number of the hydrogen atom comes closest to giving a 61-nm-diameter electron orbit? b. What are the electron’s speed

Question

a. What quantum number of the hydrogen atom comes closest to giving a 61-nm-diameter electron orbit?
b. What are the electron’s speed and energy in this state?

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Mộc Miên 5 years 2021-08-06T02:12:11+00:00 1 Answers 83 views 0

Answers ( )

  1. thienthanh
    0
    2021-08-06T02:13:23+00:00
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    Answer:

    a

      n = 23

    b

      v = 87377.95 \ m/s

    Explanation:

    From the question we are told that

       The diameter is d = 61\ nm = 61 *10^{-9} \ m

       

    Generally the radius electron orbit  is mathematically represented as

          r = \frac{61 *10^{-9}}{2}

    =>   r = 3.05*10^{-8} \ m

    This radius can also be represented mathematically  as

          r = n^2 * a_o

    Here n is the quantum number and a_o is  the Bohr radius with a value

        a_o = 0.0529 *10^{-9} \ m

    So

       n = \sqrt{\frac{3.05*10^{-8}}{ 0.059*10^{-9}} }

    =>   n = 23

    Generally the angular momentum of the electron is mathematically represented as

              L = m * v * r = \frac{n * h }{2 \pi}

    Here  h is the Planck constant and the value is  h = 6.626*10^{-34} J \cdot s

              m is the mass of the electron with values m = 9.1*10^{-31} \ kg

             So

                   v = \frac{23 * 6.626*10^{-34} }{2\pi * 9.1 *10^{-31} * 3.05*10^{-8} }

                    v = 87377.95 \ m/s

           

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