Consider the Bohr energy expression (Equation 30.13) as it applies to singly ionized helium He+ (Z = 2) and an ionized atom with Z=4 and only a single electron orbiting the nucleus. This expression predicts equal electron energies for these two species for certain values of the quantum number n (the quantum number is different for each species). For quantum numbers less than or equal to 9, what is the highest energy (in electron volts) for which the helium energy level is equal to the ionized atom energy level?
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The Bohr energy expression (Equation 30.13) for the energy levels of an electron in an atom is given by:
E = -13.6 eV * (Z^2 / n^2)
where E is the energy of the electron, Z is the atomic number, and n is the principal quantum number.
For the singly ionized helium ion, He+ (Z = 2), the energy expression becomes:
E(He+) = -13.6 eV * (2^2 / n^2)
For an ion with Z = 4 and only a single electron, the energy expression becomes:
E(Z=4) = -13.6 eV * (4^2 / n^2)
We need to find the highest energy (in electron volts) for which the helium energy level is equal to the ionized atom energy level. So we need to find the value of n for which E(He+) = E(Z=4).
Setting the two energy expressions equal to each other:
-13.6 eV * (2^2 / n^2) = -13.6 eV * (4^2 / n^2)
Simplifying the equation:
2^2 / n^2 = 4^2 / n^2
4 / n^2 = 16 / n^2
Cross-multiplying:
4n^2 = 16
Dividing both sides by 4:
n^2 = 4
Taking the square root of both sides:
n = ±2
Since we are looking for the highest energy for n <= 9, the only valid solution is n = 2.
Plugging in this value into the energy expression for Helium:
E(He+) = -13.6 eV * (2^2 / 2^2)
E(He+) = -13.6 eV
Therefore, the highest energy (in electron volts) for which the helium energy level is equal to the ionized atom energy level for quantum numbers less than or equal to 9 is -13.6 eV.
To find the highest energy at which the helium energy level is equal to the ionized atom energy level for quantum numbers less than or equal to 9, we can use the Bohr energy expression.
The Bohr energy expression for the energy level of an electron orbiting a nucleus is given by:
E = -13.6 * (Z^2 / n^2) eV
Where:
E is the energy
Z is the atomic number (number of protons in the nucleus)
n is the principal quantum number
For the singly ionized helium (He+) with Z = 2, the energy expression becomes:
E1 = -13.6 * (2^2 / n^2) eV
For the ionized atom with Z = 4, the energy expression becomes:
E2 = -13.6 * (4^2 / n^2) eV
To find when the energy levels are equal, we set E1 equal to E2:
-13.6 * (2^2 / n^2) = -13.6 * (4^2 / n^2)
Simplifying the equation, we have:
4 / n^2 = 16 / n^2
Since the denominators are the same, we can eliminate them:
4 = 16
This equation is not possible, so there are no values of 'n' for which the helium energy level is equal to the ionized atom energy level for quantum numbers less than or equal to 9.
Therefore, there is no highest energy for which the helium energy level matches the ionized atom energy level for quantum numbers up to 9.