Final answer:
To calculate the energy left over after ionization in photoelectron spectroscopy using a laser with a wavelength of 121 nm, we can use the formula E = hc/λ. The energy of the laser photon is approximately 10.22 eV, and subtracting the ionization energy of tyrosine (7.5 eV) from this value gives us 2.72 eV left over after ionization.
Step-by-step explanation:
Photoelectron spectroscopy is based on the photoelectric effect, which measures the energy required to remove an electron from the highest-occupied molecular orbital and ionize the molecule. To calculate the energy left over after ionization, we need to know the energy of the laser photon.
We can use the formula E = hc/λ, where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of the laser. In this case, the wavelength is given as 121 nm, so we have E = (6.626 x 10^-34 J·s) x (3.00 x 10^8 m/s) / (121 x 10^-9 m). Converting from joules to electron volts, 1 eV = 1.602 x 10^-19 J, we find that the energy of the laser photon is approximately 10.22 eV.
Since the ionization energy of tyrosine is 7.5 eV, subtracting this value from the energy of the laser photon gives us the remaining energy: 10.22 eV - 7.5 eV = 2.72 eV left over after ionization.