A technique called photoelectron spectroscopy is used to measure the ionization

Question

A technique called photoelectron spectroscopy is used to measure the ionization energy of atoms. A sample is irradiated with ultraviolet (UV) light, and electrons are ejected from the valence shell. The kinetic energies of the ejected electrons are measured. Because the energy of the UV photon and the kinetic energy of the ejected electron are known, we can write hv=IE+1/2mv^2, in which v is the frequency of the UV light, and m and v are the mass and velocity of the electron, respectively. In one experiment the kinetic energy of the ejected electron from potassium is found to be 5.34x10^-19 J using a UV source of wavelength 162 nm. Calculate the ionization energy of potassium (in what units is it usually reported?). Answer: 420 kj/mol Please help with the steps and how they got this answer? Thank you!

Explanation / Answer

given

K.E = 0.5 mv^2 = 5.34 x 10-19 J

now

hv = h x c / lamda

given

lamda = wavelength = 162 nm = 162 x 10-9 m

so

hv = h x c / lamda = 6.626 x 10-34 x 3 x 10^8 / 162 x 10-9

hv = h x c / lamda = 1.227 x 10-18

now

IE = hv - K.E

IE = (1.227 x 10-18) - ( 5.34 x 10-19)

IE = 6.93037 x 10-19

this is per one atom

we know that

1 mole contains 6.023 x 10^23 atoms

so

IE = 6.93037 x 10-19 x 6.023 x 10^23 J/mol

IE = 417416 J/mol

IE = 417.416 x 1000 J/mol

IE = 417.416 kJ/mol

so

the ionization energy is 417.416 kJ/mol

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.