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# Answer to Question #120364 in Mechanics | Relativity for Grasya ni, bawal pinamay

Question #120364
VII. The photoelectric threshold wavelength of a tungsten surface is 272 nm. Calculate the maximum
kinetic energy of the electrons ejected from this tungsten surface by ultraviolet radiation of frequency
1.45 x 1015 Hz. Express the answer in electron volts.

Notes:
Remember that E=hf=hc/λ
Threshold wavelength means the minimum amount of wavelength to remove electrons from a surface
so in this problem, E = = hf ϕ threshold=hc/ λthreshold
where λthreshold = threshold wavelength
= work function = the minimum energy needed to remove an electron from the surface ϕ
h = Planck’s constant = 6.626 x 10 -34 J.s = 4.136 x 10 -15 eV.s
c = 3x108 m/s
f = frequency.
1
2020-06-08T10:29:45-0400

"\\frac{hc}{\\lambda}=A+KE"

"A=\\frac{hc}{\\lambda_0}=\\frac{6.62\\cdot 10^{-34}\\cdot 3\\cdot10^8}{272\\cdot10^{-9}}=7.3\\cdot10^{-19}J"

"KE=h\\nu-7.3\\cdot10^{-19}=6.62\\cdot 10^{-34}\\cdot 1.45\\cdot10^{15}-7.3\\cdot10^{-19}="

"=2.3\\cdot10^{-19}J=1.44eV"

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