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Mastering Chem help!!!?
1. What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.6×1015Hz?
1. What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.6×1015Hz?
KE= ______ J
1 Answer
- BrentLv 64 years agoFavorite Answer
The energy applied will do two things:
1. Some energy will remove the electron from the cesium atom
2. The rest of the energy will give the electron kinetic energy.
First, we apply Planck's constant to determine the total energy transferred to the electron.
E = hf = (6.626 × 10^-34 J/Hz)(1.6 × 10^15 Hz) = 1.06 × 10^-18 J
Next, the first ionization energy for cesium:
375.6 kJ/mol * (1 mol / 6.022 × 10^23 electrons) = 6.237 × 10^-19 J per electron
Finally, we subtract the ionization energy from the total energy of the UV ray to see what is left for kinetic energy:
KE = 1.06 × 10^-18 J - 6.237 × 10^-19 J = 4.36 × 10^-19 J
