The ionization energy of an atom can be measured by photoelectron spectroscopy,

Question

The ionization energy of an atom can be measured by photoelectron spectroscopy, in which light wavelength (lamda) is directed at an atom, causing an electron to be ejected. The kentic energy of the ejected electron (Ek)is measured by determining the velocity, v, since Ek=1/2mv^2. The Ei is then calculated using the relationship that the energy of the incident light equals the sum Ei + Ek.
(a) What is the ionization energy of rubidium atoms in kJ/mol if light with lamda=58.4nm produces electrons with a velocity of 2.450*10^6m/s? ( the mass ofan electron is 9.109*10^-31kg)
(b) What is the ionization energy of potassium in kJ/mol if light with lamda=142nm produces electrons with a velocity of 1.240*10^6m/s?

Explanation / Answer

Total energy = work function (ionization energy) + kinetic energy

Energy= h (planck's constant, 6.626 x 10^-34 m^2 kg/ s) * c (speed of light) / lamda

SO,

A) (6.626 X 10^-34) * (3.00 * 10^8) / (58.4 * 10^-9) = Ei + (1/2) (9.109*10^-31)(2.450*10^6)^2)

so Ei = 2.446 *10^-18 Joules

** make sure you keep your units straight (convert nm to m for wavelength)

B) (6.626 X 10^-34) * (3.00 * 10^8) / (142 * 10^-9) = Ei + (1/2) (9.109*10^-31)(1.240*10^6)^2)

so Ei= 6.996 *10^-19 Joules

Hope this helps!

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