一到英文的物理题Potassium is illuminated with UV light of wavelength 2500 Angstroms.If theWork function of potassium is 2.21 eV,what is the maximum kinetic energy of theemitted electrons?[1eV = 1.6 x 10‐19 Joules]
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一到英文的物理题Potassium is illuminated with UV light of wavelength 2500 Angstroms.If theWork function of potassium is 2.21 eV,what is the maximum kinetic energy of theemitted electrons?[1eV = 1.6 x 10‐19 Joules]
一到英文的物理题
Potassium is illuminated with UV light of wavelength 2500 Angstroms.If the
Work function of potassium is 2.21 eV,what is the maximum kinetic energy of the
emitted electrons?[1eV = 1.6 x 10‐19 Joules]
一到英文的物理题Potassium is illuminated with UV light of wavelength 2500 Angstroms.If theWork function of potassium is 2.21 eV,what is the maximum kinetic energy of theemitted electrons?[1eV = 1.6 x 10‐19 Joules]
The maximum kinetic energy Ekmax is :
ħf - ф = Ekmax
Where
ħ = plancks constant = 6.626x10^-34 J s
f = frenquency of the UV light = 1 / wavelength = 1 / (2500 X 10^-10m) =4X10^6 /s
ф = work function of potassium = 2.21 eV
1 eV = 1.6x10^-19 J
2.21eV = 3.536x10^-19 J
Ekmax = (6.626x10^-34 J s) x (4x10^6 / s) - 3.536x10^-19 J