114k views
4 votes
For carbon, calculate the energy when an electron falls from n=3 to n=2, then from n=2 to n=1. Then add these for the total energy released in the process.

a) ( E_total = 13.6 , eV )
b) ( E_total = 8.2 , eV )
c) ( E_total = 10.4 , eV )
d) ( E_total = 6.8 , eV )

User David Gras
by
8.1k points

1 Answer

3 votes

Final answer:

The total energy released when an electron transitions from n=3 to n=2 and then from n=2 to n=1 can be calculated using the formula En = -13.6 eV/n², yielding a total energy release of 12.09 eV.

Step-by-step explanation:

To calculate the energy when an electron falls from one energy level to another in a carbon atom, the energy of each level for carbon can be represented similarly to the hydrogen case (albeit incorrectly for the carbon's actual energy levels). Using the formula for hydrogen-like atoms, En = -13.6 eV/n², we can find the energy of the n=3 and n=2 levels. For n=3, the energy is E3 = -13.6 eV/3² = -1.51 eV, and for n=2, it's E2 = -13.6 eV/2² = -3.4 eV.

For the transition from n=3 to n=2, the energy released is ΔE = E2 - E3 = -3.4 eV - (-1.51 eV) = 1.89 eV. Similarly, for the transition from n=2 to n=1, the energy released is ΔE = E1 - E2, where E1 is the energy of the first level. Since E1 = -13.6 eV (the energy of the electron in n=1 for hydrogen-like atoms), the energy released for this transition is ΔE = -13.6 eV - (-3.4 eV) = 10.2 eV.

Adding the energy changes together: E_total = 1.89 eV + 10.2 eV = 12.09 eV. This is not an exact value for carbon, as carbon is not a hydrogen-like atom and this formula is based on hydrogen's spectral lines, but it gives a hypothetical scenario. There are differences in the energy levels due to the multielectron nature of carbon and its corresponding different shielding and electron-electron repulsions.

To address part (b), you would calculate the energy released in a direct fall from n=3 to n=1 for carbon, using the formula ΔE = E1 - E3. Finally, addressing part (c), the energies in part (a) and part (b) are not the same due to the difference in the initial and final states of the transitions.

User Ganesh Ghalame
by
7.7k points

No related questions found