A Helium atom contains one electron in the n = 6 energy orbital. Calculate the energy of the electron. Give your answer in SI units with the correct sign.
Question
A Helium atom contains one electron in the n = 6 energy orbital. Calculate the energy of the electron. Give your answer in SI units with the correct sign.
Solution
The energy of an electron in an atom is given by the formula:
E = -13.6 * Z^2 / n^2 eV
where:
- E is the energy of the electron,
- Z is the atomic number of the atom (for helium, Z = 2),
- n is the principal quantum number (in this case, n = 6).
Substituting the values into the formula, we get:
E = -13.6 * 2^2 / 6^2 eV E = -13.6 * 4 / 36 eV E = -1.51 eV
However, the question asks for the answer in SI units. The conversion factor from electron volts to joules is 1 eV = 1.6 x 10^-19 J. Therefore, the energy in joules is:
E = -1.51 eV * 1.6 x 10^-19 J/eV E = -2.42 x 10^-19 J
The negative sign indicates that the electron is bound to the nucleus.
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