Chapter 11, Problem 11D.5E

(a)

Interpretation Introduction

Interpretation:

Energy of the ejected electron has to be determined.

Concept Introduction:

Photoelectron spectroscopy is based on photoelectric effect where electrons are ejected from a molecule when a beam of ultraviolet radiation or X-ray hits on it.

Ionization energy of the molecule (I) can be calculated as follows,

  $\begin{array}{c}h\nu \text{ = I + }\frac{1}{2}{m}_e{\text{υ}}^2\\ \\ where,\left(h\nu -\text{I }\right)\text{= Kinetic energy of the ejected electron}\\ \frac{1}{2}{m}_e{\text{υ}}^2=\text{Kinetic energy of an electron travelling at a speed, υ}\end{array}$

Wavelength and frequency are inversely proportional to each other and can be given as,

  $\begin{array}{c}\text{ν}\text{=}\frac{\text{c}}{\text{λ}}\\ \text{where, ν}=\text{frequency}\\ \text{λ}=\text{wavelength}\\ \text{c = speed of light}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Speed of the ejected electron has to be determined.

Concept Introduction:

Photoelectron spectroscopy is based on photoelectric effect where electrons are ejected from a molecule when a beam of ultraviolet radiation or X-ray hits on it.

Ionization energy of the molecule (I) can be calculated as follows,

  $\begin{array}{c}h\nu \text{ = I + }\frac{1}{2}{m}_e{\text{υ}}^2\\ \\ \left(h\nu -I\right)\text{ = }\frac{1}{2}{m}_e{\text{υ}}^2\\ \\ where,\left(h\nu -\text{I }\right)\text{= Kinetic energy of the ejected electron}\\ \frac{1}{2}{m}_e{\text{υ}}^2=\text{Kinetic energy of an electron travelling at a speed, υ}\end{array}$