What is the reaction that corresponds to the first ionization energy of rubidium

Rb(g) → Rb⁺(g) + e⁻    I₁ = 1st ionization energy

Explanation:

Here we have

Rb(g) → Rb⁺(g) + e⁻

The equation is of the form

X → X⁺ + e⁻

An atom that is made up of more than one electron requires a steadily  increasing amount of energy to remove each successive electron. The ionization energy I₁ to remove the first electron is the first ionization energy, which can be defined as follows

E(g) → E⁺(g)+ e⁻    I₁ = 1st ionization energy

Similarly, the second ionization energy  is given by

E⁺(g) → E²⁺(g) + e⁻    I₂ = 2nd ionization energy and so forth.

the answer is false

explanation:

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.