Titanium (work function= 6.94*(10^-19) Joules) and Silicon (work function=7.77*(10^-19) Joules) surfaces are irradiated with UV radiation with a wavelength of 241 nanometers.
What is the wavelength of the electrons emitted by the Titanium surface?
2007-11-02 08:53:23 · 2 answers · asked by majoraganondorf 1 in Science & Mathematics ➔ Physics
hc = 1240eV*nm, thus the light has energy 5.145eV, the Ti work function is 4.33eV. Thus electrons escaping the surface have a maximum energy of 0.815eV. Since wavelength is h/p, and energy is p^2/(2m), wavelength = h(2mE)^(-1/2) = 1.36nm, meaning 1.36nm wavelengths or greater (lower energy) are emitted.
2007-11-02 09:10:12 · answer #1 · answered by supastremph 6 · 0⤊ 1⤋
photoelectric effect
e=hc/wavelength
=energy of the incident photon
this minus work function gives energy of electron
wavelength of electron is found from the energy of it, and de broglie theory...
2007-11-02 09:14:14 · answer #2 · answered by Anonymous · 0⤊ 0⤋