Question

One way to measure ionization energies is ultraviolet photoelectron spectroscopy (UPS, or just PES), a technique based on the photoelectric effect. In PES, monochromatic light is directed onto a sample, causing electrons to be emitted. The kinetic energy of the emitted electrons is measured. The difference between the energy of the photons and the kinetic energy of the electrons corresponds to the energy needed to remove the electrons (that is, the ionization energy). Suppose that a PES experiment is performed in which mercury vapor is irradiated with ultraviolet light of wavelength 58.4 nm.

(a) What is the energy of a photon of this light in eV?
(b) Write an equation that shows the process corresponding to the first ionization energy of Hg.
(c) 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?

Answer 1

Step-by-step explanation:

From the given information:

The energy of photons can be determined by using the formula:

`E = (hc)/(\lambda)`

where;

planck's constant (h) =
`6.63 * 10^ {-34}`

speed oflight (c) =
`3.0 * 10^8 \ m/s`

wavelength λ = 58.4 nm

`E = (6.63 * 10^(-34) \ J.s * 3.0 * 10^8 \ m/s)/(58.4 * 10^(-9 ) \ m)`

`E =0.34 * 10^(-17) \ J`

`E = 3.40 * 10^(-18 ) \ J`

To convert the energy of photon to (eV), we have:

`1 eV = 1.602 * 10^(-19) \ J`

Hence

`3.40 * 10^(-18 ) \ J = (1 eV)/(1.602 * 10^(-19 ) \ J )* 3.40 * 10^(-18 ) \ J`

`E = 2.12 * 10 \ eV`

E = 21.2 eV

b)

The equation that illustrates the process relating to the first ionization is:

`Hg_((g)) \to Hg^+ _((g)) + e^-`

c)

The 1st ionization energy (I.E) of Hg can be calculated as follows:

Recall that:

I.E = Initial energy - Kinetic Energy

I₁ (eV) = 21.2 eV - 10.75 eV

I₁ (eV) = 10.45 eV

Since ;

`1 eV = 1.602 * 10^(-19) \ J`

∴

`10.45 \ eV = (1.602 * 10^(-19 ) \ J )/( 1 \ eV)* 10.45 \ eV`

Hence; the 1st ionization energy of Hg atom =
`1.67 * 10^(-18) \ J`

`1.67 * 10^(-21) \ kJ`

Finally;

`I_1 \ of \ the \ Hg (kJ/mol) = (1.67 * 10^(-21 \ kJ) * 6.02 * 10^(23) \ Hg \ atom )/(1 \ Kg \ atom )`

`\mathbf{= 1.005 * 10^3 \ kJ/mol}`