Using the Rydberg equation, calculate the energy for the following electronic transitions in a hydrogen atom and label each as an absorption or an emission. When calculating, watch your signs! n = 3 → n = 1

a. Energy calculation
b. absorption or emission?

-1.94  * 10^-18 J

Since the electron moved from a higher to a lower energy level (n = 3 → n = 1) it is an emission.

Explanation:

From Rydberg equation;

E = -RH(1/n^2final - 1/n^2initial)

For a transition from  n = 3 → n = 1

RH = 2.18 * 10^-18 J

E = -(2.18 * 10^-18) (1/1^2 - (1/3^2)

E = -(2.18 * 10^-18) (1-1/9)

E= -(2.18 * 10^-18) (8/9)

E = -1.94  * 10^-18 J

answer by yourhope:

how do models compare to real-world objects?

d. most models represent a real-world object in a simpler way.

: )

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

if you need further assistance, let me know: )

~ trueboss72