🤓 Based on our data, we think this question is relevant for Professor Johnson's class at UCSD.

The first ionization energy of carbon is 1.81 aJ. Assuming an ionization efficiency of 22.35%, how many photons of the lowest possible frequency are required to ionize a sample of carbon that contains 5.32 x 10^{18} atoms?

a) 2.38 x 10^{19} photons

b) 4.20 x 10^{-20} photons

c) 2.73 x 10^{15} photons

d) 9.11 x 10^{6} photons

We’re being asked to calculate how many photons of the lowest possible frequency are required to ionize a sample of carbon. The first ionization energy of carbon is 1.81 aJ.

*Ionization energy (IE) **is the energy required to remove an electron from **a gaseous atom **or ion. Let’s calculate the **ionization** energy in J.*

**1 aJ = 10 ^{-18} J**

$\mathit{I}\mathit{E}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{81}\mathbf{}\overline{)\mathbf{a}\mathbf{J}}\mathbf{}\mathbf{\times}\mathbf{}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{18}}\mathbf{}\mathbf{J}}{\mathbf{1}\mathbf{}\overline{)\mathbf{a}\mathbf{J}}}$

*IE = 1.81x10 ^{-18} J/atom*

The *ionization efficiency is 22.35%*. Calculate the energy for 22.35% ionization efficiency:

**% ionization efficiency = 22.35 %**

**Total energy = 1.81x10 ^{-18} J/atom**

Photoelectric Effect

Photoelectric Effect

Photoelectric Effect

Photoelectric Effect