In order to determine the reaction that corresponds to the first ionization energy of rubidium, we must first understand what ionization energy is. Ionization energy is the amount of energy required to remove an electron from an atom or molecule. For rubidium, the first ionization energy is 4.177 eV. This means that it takes 4.177 eV to remove the first electron from a rubidium atom.
The reaction that corresponds to the first ionization energy of rubidium can be represented as follows:
Rb + energy -> Rb+ + e-
In this reaction, rubidium (Rb) absorbs energy and produces a rubidium ion (Rb+) and an electron (e-). This reaction requires 4.177 eV of energy, which is the first ionization energy of rubidium.
Other related questions:
Q: What is RB first ionization energy?
A: The first ionization energy of RB is the amount of energy required to remove the outermost electron from a neutral RB atom.
Q: What is the ionization energy of Rb+?
A: The ionization energy of Rb+ is 4.18 eV.
Q: Does SR or RB have a higher ionization energy?
A: The answer to this question depends on the specific elements involved. Generally speaking, elements in the same group tend to have similar ionization energies. However, there are some exceptions. For example, sulfur (S) has a higher ionization energy than tellurium (Te).