(c) The kinetic energy of the emitted electrons is measured to be 15.2 eV. What is the first ionization energy of Hg in kJ/mol?
Incomplete question
The first ionization energy (IE) is the energy required to remove one electron from an atom or ion in its gaseous state. To determine the first ionization energy of Hg, we need to convert the given value of kinetic energy from electron volts (eV) to joules (J).
1 eV = 1.602 × 10^-19 J
Given that the kinetic energy of the emitted electrons is 15.2 eV, we can convert it to joules:
15.2 eV × 1.602 × 10^-19 J/eV = 2.44 × 10^-18 J
Now, we need to convert the energy from joules to kilojoules (kJ):
1 J = 1 × 10^-3 kJ
2.44 × 10^-18 J × 1 × 10^-3 kJ/J = 2.44 × 10^-21 kJ
The next step is to calculate the number of moles of Hg present. We can use Avogadro's number, which states that 1 mole of any substance contains 6.022 × 10^23 particles.
Finally, we divide the energy in kilojoules by the number of moles to find the first ionization energy of Hg in kJ/mol:
First ionization energy = (2.44 × 10^-21 kJ) / (6.022 × 10^23 mol^-1)
Calculating this division will give us the final answer for the first ionization energy of Hg in kJ/mol.