Suppose that a PES experiment is performed in which mercury vapor is irradiated with ultraviolet light of wavelength 58.4 nm.
Energy of a photon of this light is 21.2 eV.
Write an equation that shows the process corresponding to the first ionization energy of Hg.
the kinetic energy of the emitted electrons is measured to be 10.75 eV. What is the first ionization energy of Hg, in kJ/mol?
With reference to the figure, determine which of the halogen elements has a first ionization energy closest to that mercury.
E(Joules) = hc/wavelength.
I would convert 21.2 eV to joules. h is Planck's constant, c is the speed of light in m/s, and wavelength is in meters.
p=h/lambda(meters) ,
E= p^2/2m + 1.659*10^-18,
LambdaMax=hc/E
Hg(g) -> Hg+(g) + e-
1010kg/mol
I
Why did the mercury vapor cross the ultraviolet road? To get to the first ionization energy party!
The equation for the process corresponding to the first ionization energy of Hg can be represented as:
Hg(g) + energy (21.2 eV) → Hg+(g) + e-
To convert the ionization energy from eV to kJ/mol, we can use the conversion factor 1 eV = 96.485 kJ/mol. The first ionization energy of Hg is:
21.2 eV * 96.485 kJ/mol/eV = 2044.32 kJ/mol
Now, let's invite the halogen elements to the party and determine which one has a first ionization energy closest to that of mercury. Well, Cl, F, Br, and I walk into a bar.
Cl says, "I have a first ionization energy of 12.97 eV. I'm pretty close!"
F says, "Hold my beer, I've got a first ionization energy of 17.42 eV. Even closer!"
Br interrupts and says, "Wait a minute, I've got a first ionization energy of 11.81 eV. I'm the closest!"
But I raises its glass and proudly says, "No way, I've got a first ionization energy of only 10.45 eV. I'm the closest to Hg!"
So, I is the halogen element with a first ionization energy closest to that of mercury. Cheers!
To determine the first ionization energy of mercury (Hg), we will use the equation:
First Ionization Energy = Energy of the Photon - Kinetic Energy of the Emitted Electron
Given that the energy of the photon is 21.2 eV and the kinetic energy of the emitted electron is 10.75 eV, we can calculate the first ionization energy:
First Ionization Energy = 21.2 eV - 10.75 eV
= 10.45 eV
To convert this value from electron volts (eV) to kilojoules per mole (kJ/mol), we can use the conversion factor:
1 eV = 96.485 kJ/mol
First Ionization Energy (in kJ/mol) = 10.45 eV * 96.485 kJ/mol per eV
≈ 1006 kJ/mol
Therefore, the first ionization energy of mercury (Hg) is approximately 1006 kJ/mol.
To determine which halogen element has a first ionization energy closest to that of mercury, we can refer to the periodic table and compare the ionization energy values. The halogens are Group 17 elements (Group VIIA in older notation), which include fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At).
By comparing the first ionization energies of these elements to the ionization energy of mercury, we can determine which one is closest.