If the outermost electron in an atom is excited to a very high energy state, its orbit is far beyond that of the other electrons. To a good

Question

If the outermost electron in an atom is excited to a very high energy state, its orbit is far beyond that of the other electrons. To a good approximation, we can think of the electron as orbiting a compact core with a charge equal to the charge of a single proton. The outer electron in such a Rydberg atom thus has energy levels corresponding to those of hydrogen.
Sodium is a common element for such studies. How does the radius you calculated in part A compare to the approximately 0.20 nm radius of a typical sodium atom?
r100/rNa = _______.

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Trung Dũng 6 months 2021-07-21T02:07:32+00:00 1 Answers 12 views 0

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    2021-07-21T02:08:58+00:00

    Answer:

    the calculated ratio to the radius of the sodium r_{100 / r_{Na is 2645.0

    Explanation:

    Given the data in the question;

    the calculated ratio to the radius of the sodium = r_{100 / r_{Na

    so from here we can write the number of energy states as 100

    The number of energy states; n = 100

    A;

    We know that the radius of the sodium atom is;

    r_n = n²α₀

    Now, the value of the Bohr radius; α₀ = 5.29 × 10⁻¹¹ m

    so lets determine the radius of the sodium atom; by substituting in our values;

    r_{100 = (100)² × (5.29 × 10⁻¹¹ m )

    r_{100 = 5.29 × 10⁻⁷ m

    B

    given that, the theoretical value of the radius of the sodium is;

    r_{Na = 0.2 nm = 2 × 10⁻¹⁰ m

    so we calculate the ratio of the radii of the sodium;

    r_{100 / r_{Na = ( 5.29 × 10⁻⁷ m ) / ( 2 × 10⁻¹⁰ m )

    r_{100 / r_{Na = 2645.0

    Therefore, the calculated ratio to the radius of the sodium r_{100 / r_{Na is 2645.0

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