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

Question

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.

Answer 1

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

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

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

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