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 = .

the calculated ratio to the radius of the sodium / is 2645.0

Explanation:

Given the data in the question;

the calculated ratio to the radius of the sodium = /

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;

= 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;

= (100)² × (5.29 × 10⁻¹¹ m )

= 5.29 × 10⁻⁷ m

B

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

= 0.2 nm = 2 × 10⁻¹⁰ m

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

/ = ( 5.29 × 10⁻⁷ m ) / ( 2 × 10⁻¹⁰ m )

/ = 2645.0

Therefore, the calculated ratio to the radius of the sodium / is 2645.0

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.