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

Answer:

I = 1010 kJ/mol

Explanation:

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

Energy of the incident photon

$E = \dfrac{hc}{\lambda}$

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

c is speed of light = 3 x 10⁸ m/s

λ = wavelength

$E = \dfrac{6.626 \times 10^{-34}\times 3 \times 10^8}{58.4 \times 10^{-9}}$

E = 3.40 x 10⁻¹⁸ J

Energy in eV

$E = \dfrac{3.40 \times 10^{-18}}{1.6 \times 10^{-19}}$

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