Question

What minimum frequency of light is required to ionize boron?

Answer 1

Answer:

frequency  = 8.22 x 10¹⁴ s⁻¹

Explanation:

An electron's positional potential energy while in a given principle quantum energy level is given by Eₙ = - A/n² and A = constant = 2.18 x 10⁻¹⁸j. So to remove an electron from the valence level of Boron (₅B), energy need be added to promote the electron from n = 2 to n = ∞. That is, ΔE(ionization) = E(n=∞) - E(n=2) = (-A/(∞)²) - (-A/(2)²) = [2.18 x 10⁻¹⁸j/4] joules = 5.45 x 10⁻¹⁹ joules.

The frequency (f) of the wave ionization energy can then be determined from the expression ΔE(izn) = h·f; h = Planck's Constant = 6.63 x 10⁻³⁴j·s. That is:

ΔE(izn) = h·f => f = ΔE(izn)/h = 5.45 x 10⁻¹⁹ j/6.63 x 10⁻³⁴ j·s = 8.22 x 10¹⁴ s⁻¹