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Ksivusya [100]
2 years ago
6

When an excited electron spontaneously emits a photon, which energy transformation takes place

Chemistry
2 answers:
kompoz [17]2 years ago
7 0

When an excited electron spontaneously emits a photon, the energy transformation from potential to kinetic energy takes place.

If an atom is in an excited state means full of energy, it goes from higher energy level to lower energy level. The electron releasing energy in the form of a photon, which is emitted in a random direction.

When the electron returns to a low energy state, it releases the potential energy in the form of kinetic energy so we can conclude that the energy transformation from potential to kinetic energy takes place.

Learn more about excited electron here: brainly.com/question/81112

Learn more: brainly.com/question/25976890

notka56 [123]2 years ago
4 0

Given what we know, we can confirm that when an excited electron spontaneously emits a photon, the energy released is electromagnetic energy.

<h3>What is a Photon and what energy does it release when being emitted?</h3>
  • A photon is a particle.
  • This means that it is one of the smallest forms of matter that we can study.
  • Photons form electromagnetic fields.
  • Therefore, when being emitted by an electron, photons release electromagnetic energy.

Therefore, we can confirm that when an excited electron spontaneously emits a photon, the energy released is electromagnetic energy due to the properties of the photon being emitted.

To learn more about photons visit:

brainly.com/question/24309591?referrer=searchResults

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We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants
NeX [460]

Explanation:

Reaction equation is as follows.

      Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)

Here, 1 mole of Na_{2}SO_{3} produces 2 moles of cations.

[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58

                                  = 1.16 M

[SO^{2-}_{3}] = [Na_{2}SO_{3}] = 0.58 M

The sulphite anion will act as a base and react with H_{2}O to form HSO^{-}_{3} and OH^{-}.

As,     K_{b} = \frac{K_{w}}{K_{a_{2}}}

                       = \frac{10^{-14}}{6.3 \times 10^{-8}}

                       = 1.59 \times 10^{-7}

According to the ICE table for the given reaction,

          SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}

Initial:        0.58             0              0

Change:     -x               +x             +x

Equilibrium: 0.58 - x     x               x

So,

        K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}

 1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}

        x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)

                x = 0.0003 M

So,   x = [HSO^{-}_{3}] = [OH^{-}] = 0.0003 M

[SO^{2-}_{3}] = 0.58 - 0.0003

                     = 0.579 M

Now, we will use [HSO^{-}_{3}] = 0.0003 M

The reaction will be as follows.

              HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}

Initial:   0.0003

Equilibrium:  0.0003 - x        x             x

              K_{b} = \frac{x^{2}}{0.0003 - x}

        K_{b} = \frac{K_{w}}{K_{a_{1}}}

                      = \frac{10^{-14}}{1.4 \times 10^{-2}}

                      = 7.14 \times 10^{-13}

Therefore,  7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}

As,  x <<<< 0.0003. So, we can neglect x.

Therefore,  x^{2} = 7.14 \times 10^{-13} \times 0.0003

                              = 0.00214 \times 10^{-13}

                     x = 0.0146 \times 10^{-6}

x = [OH^{-}] = [H_{2}SO_{3}] = 1.46 \times 10^{-8}

    [H^{+}] = \frac{10^{-14}}{[OH^{-}]}

                = \frac{10^{-14}}{0.0003}

                = 3.33 \times 10^{-11} M

Thus, we can conclude that the concentration of spectator ion is 3.33 \times 10^{-11} M.

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