Question

An xray photon of wavelength 0.989 nm strikes a surface. The emitted electron has a kinetic energy of 969eV. What is the binding energy of the electron in kJ/mol?

Answer 1

Answer on Question #40615-Chemistry-Inorganic Chemistry

Question

An X-ray photon of wavelength 0.989 nm strikes a surface. The emitted electron has a kinetic energy of 969 eV. What is the binding energy of the electron in kJ/mol?

Solution

The electron binding energy of each of the emitted electrons can be determined by using an equation that is based on the work of Ernest Rutherford:

$$E_{binding} = E_{photon} - (E_{kinetic} + \varphi)$$

where $E_{binding}$ is the binding energy of the electron, $E_{photon}$ is the energy of the X-ray photons being used, $E_{kinetic}$ is the kinetic energy of the electron as measured by the instrument and $\varphi$ is the work function of the spectrometer.

It is given that $E_{kinetic} = 969 \, \text{eV} = 969 \cdot 1.60 \cdot 10^{-19} = 1.55 \cdot 10^{-16} \, \text{J}$

Since the work function $\varphi$ is not specified in the task it can be neglected ($\varphi = 0$).

Energy of the photon is

$$E_{photon} = h\nu = \frac{hc}{\lambda},$$

where $h$ – Plank constant ($h = 6.63 \cdot 10^{-34} \, \text{J} \cdot \text{s}$), $c$ – speed of light ($c = 3.00 \cdot 10^{8} \, \text{m} \cdot \text{s}^{-1}$), $\lambda$ – wavelength of the photon. In given case $\lambda = 0.989 \, \text{nm} = 9.89 \cdot 10^{-10} \, \text{m}$

$$E_{photon} = \frac{hc}{\lambda} = \frac{6.63 \cdot 10^{-34} \cdot 3.00 \cdot 10^{8}}{9.89 \cdot 10^{-10}} = 2.01 \cdot 10^{-16} \, \text{J}$$

The binding energy:

$$E_{binding} = E_{photon} - E_{kinetic} = 2.01 \cdot 10^{-16} - 1.55 \cdot 10^{-16} = 4.60 \cdot 10^{-17} \, \text{J} = 4.60 \cdot 10^{-20} \, \text{kJ}$$

To convert the value to kJ/mol we should multiply it by Avogadro constant ($N_A = 6.02 \cdot 10^{23} \, \text{mol}^{-1}$)

$$E_{binding} = 4.60 \cdot 10^{-20} \, \text{kJ} \cdot 6.02 \cdot 10^{23} \, \text{mol}^{-1} = 27692 \, \text{kJ/mol}$$

Answer: 27692 kJ/mol