A technique called photo electron 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 \( h v=I E+\frac{1}{2} m u^{2} \) in which \(\nu\) is the frequency of the UV light, and \(m\) and \(u\) 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 \(5.34 \times 10^{-19} \mathrm{~J}\) using a UV source of wavelength \(162 \mathrm{nm}\). Calculate the ionization energy of potassium. How can you be sure that this ionization energy corresponds to the electron in the valence shell (that is, the most loosely held electron)?

Step by step solution

01

Calculate the frequency of the UV light

The frequency (\(\nu\)) of the UV light can be calculated from the wavelength (\(\lambda\)) using the formula \(\nu = \frac{c}{\lambda}\), where \(c\) is the speed of light. Given that the speed of light \(c = 3 \times 10^8 \, m/s\) and the wavelength \(\lambda = 162 \times 10^{-9} \, m\) (162 nm), the frequency can be calculated as follows: \(\nu = \frac{3 \times 10^8 \, m/s}{162 \times 10^{-9} \, m} = 1.85 \times 10^{15}\, Hz\).

02

Calculate the energy of the incident UV photon

The energy of the incident UV photon can be calculated using the formula \(E = h\nu\), where \(h\) is Planck’s constant (\(6.63 \times 10^{-34} \, J \cdot s\)). The energy of the photon is: \(E = 6.63 \times 10^{-34}\, J \cdot s \times 1.85 \times 10^{15}\, Hz = 1.23 \times 10^{-18}\, J\).

03

Calculate the ionization energy

The ionization energy can be calculated by rearranging the given equation to: \(IE = h\nu - \frac{1}{2} mu^2\). Here, \(m\) and \(u\) are the mass and velocity of the electron, respectively, which can be replaced with the kinetic energy term: \(5.34 \times 10^{-19} \, J\). Now insert the calculated photon energy and given kinetic energy into this formula to get the ionization energy: \(IE = 1.23 \times 10^{-18}\, J - 5.34 \times 10^{-19}\, J = 6.96 \times 10^{-19}\, J\).

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