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 = _______.

Answer 1

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