Respuesta :
Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V
Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:
h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.
In order to calculate the cutoff wavelength we have to consider that Ek=0
in this case h*ν=W
(h*c)/λ=4.52 eV
λ= (h*c)/4.52 eV
λ= (1240 eV*nm)/(4.52 eV)=274.34 nm
From this h*ν = Ek+W; we can calculate the kinetic energy for a radiation wavelength of 198 nm
then we have
(h*c)/(λ)-W= Ek
Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV
Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this acts to slow down the ejected electrons from the catode.
The cutoff wavelength for tungsten, the maximum kinetic energy of the electrons and the stopping potential are all mathematically given as
- w=274.34 nm
- Ek=1.74 eV
- Ek'=1.74 eV
What are the cutoff wavelength for tungsten, the maximum kinetic energy of the electrons and the stopping potential ?
Question Parameter(s):
The work function for tungsten metal is 4.52eV
the electrons when radiation of wavelength 198nm is used
Generally, the equation for the cutoff wavelength is mathematically given as
h*ν=W
Therefore
(h*c)/w=4.52
eV w= (h*c)/4.52
w= (1240 )/(4.52 eV)
w=274.34 nm
b)
h*ν = Ek+W
Ek=(h*c)/(w)-W
Ek=(1240 eV*nmnm)/(198 )-4.52 eV
Ek=1.74 eV
c)
In conclusion, The spoting potential eqauls the motion potential, therefore
Ek'=Ek
Ek'=1.74 eV
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