Question

The kinetic energy of the emitted electrons is measured to be 10.75 ev. What is the first ionization energy of hg

Answer 1

I = 1010 kJ/mol

Step-by-step explanation:

In PES experiment mercury vapor is irradiated with ultraviolet light of wavelength 58.4 nm.

Energy of the incident photon

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

h is planks constant = 6.626 × 10⁻³⁴ J.s

c is speed of light = 3 x 10⁸ m/s

λ = wavelength

`E = (6.626 * 10^(-34)* 3 * 10^8)/(58.4 * 10^(-9))`

E = 3.40 x 10⁻¹⁸ J

Energy in eV

`E = (3.40 * 10^(-18))/(1.6 * 10^(-19))`

E = 21.2 eV

first ionization energy of mercury is the difference in the kinetic energy of the ejected electron to the energy of the incident photon.

I = 21.2 eV - 10.75 eV

I = 10.45 eV

I = 10.45 x 1.6 x 10⁻¹⁹

I = 1.67 x 10⁻¹⁸ J

now,

I = (1.67 x 10⁻¹⁸ x 6.022 x 10²³) J/mol

I = 10.10 x 10⁵ J/mol

I = 1010 kJ/mol