Find the minimum energy needed to eject electrons from a metal with a threshold frequency of 3.94 × 1014 s–1.?
The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ν0. Find the minimum energy needed to eject electrons from a metal with a threshold frequency of 3.94 × 1014 s–1.
With what kinetic energy will electrons be ejected when this metal is exposed to light with a wavelength of λ = 275 nm?
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For the first one use E=hf.
E=6.62e-34 x 3.94e14
E=2.61e-19 J
For the second one use v=fl
3e+8=f x 275e-9
f=3e+8 / 275e-9
f=1.09e+15
Now use E=hf again
E=hf
E=6.62e-34 x 1.09e+15
E=7.22e-19
Now use E=E1-E2
E=7.22e-19 - 2.61e-19
E=4.61e-19 J
E=2.88eV...
Metal Threshold
use E=h*v..the place h=plank const =6.63E-34 E=abt 3E-19J=a million.873eV then use v=c/L to get frequ=3E8/235E-9=a million.277E15/s then its E=hv back..giving E=8.464E-19J..subtract paintings function (3E-19J)..leaves 5.474E-19J=3.42eV..ie the capability the photoelectron has left after it escapes the metallic floor fairly that's E(photon)=Wo+Ek(electron)..the place Wo=paintings function or the aptitude capability of the electron interior the metallic lattice
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