Question

When the hydrogen atom makes the transition from the n=2 to the n=1 energy level, it emits a photon. This photon can be absorbed by to a singly ionized helium atom (He+) in its second energy level (ninitial=2), causing it to move to a higher energy level. To what final energy level number nfinal can the photon promote the helium ion?
Note that He+ has two protons in its nucleus and a single electron.

Answer 1

Answer:

$n_{fn}$= 4

Explanation:

To solve this exercise we will use Bohr's atomic model

               $E_{n}$ = - 13.606 / n²     [eV]

The transition from level n = 2 to level n = 1 is valid

               $E_{21}$ = - 13.606 [¼ -1/1]

               $E_{21}$ = 10.2045 eV

Bohr's model for atoms with only one electron is

               $E_{n}$ = -13.606 Z² / n²

Where Z is the atomic number of the atom.

In this case the helium atom has an atomic number of Z = 2 from the level n₀ = 2 let's look up to what level it reaches

         ΔE = -13.606 [4 /  $n_{fn}$² - 4/4]

         4 / $n_{fn}$² = -ΔE / 13.606 + 1

         4 / $n_{fn}$² = -10.2045 / 13.606 +1 = -0.75 +1

         4 / $n_{fn}$² = 0.25

        $n_{fn}$ = √ 4 / 0.25

        $n_{fn}$= 4