Einstein's Equation: Calculate Frequency of Proton

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The discussion revolves around calculating the frequency of a proton with a given kinetic energy and work function, but participants identify that "proton" is likely a typo for "photon." The relevant equation, E = hf - Wo, is correctly applied to find frequency, with the formula adjusted to f = (E + Wo)/h. There is confusion regarding the use of electron volts (eV) and the conversion factor to joules, which is clarified as necessary for the calculations. The consensus is that the problem statement needs correction, specifically replacing "proton" with "photon" and ensuring proper units are used. Overall, the focus is on resolving the typographical errors and correctly applying the physics equations.
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Homework Statement


Calculte the frequency of the proton with a kinetic energy of 1.97*10-16ev if the work function of the metal is 1.33*10-16
(IeV=1.6*10-19J, h=6.60*10-34Js



Homework Equations



E = hf-hfo=hf-Wo

Where E= maximum kinetic energy

h= Plancks's constant
Wo=hfo=work function

f=frequency of the incidented light

The Attempt at a Solution



from E = hf-Wo, we have

f = E+Wo/h
f = 1.97*10-16+1.33*10-16/6.60*10-34

What about the IeV=1.6*10-19J?

Where can I subtitute that one into?
 
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The problem's statement does not make sense. Einstein's equation, which is presumably the one dealing with photoelectric emission, is about electrons and photons, not protons.

What is really given and what is required to determine?
 
voko said:
The problem's statement does not make sense. Einstein's equation, which is presumably the one dealing with photoelectric emission, is about electrons and photons, not protons.

What is really given and what is required to determine?

The problem says:

Calculte the frequency of the proton with a kinetic energy of 1.97*10-16ev if the work function of the metal is 1.33*10-16
(IeV=1.6*10-19J, h=6.60*10-34Js
 
chikis said:
Calculte the frequency of the proton with a kinetic energy of 1.97*10-16ev if the work function of the metal is 1.33*10-16
Fairly clearly 'proton' is a typo for 'photon'
(IeV=1.6*10-19J, h=6.60*10-34Js
Another typo: should be '1ev'
from E = hf-Wo, we have
f = E+Wo/h
I hope you mean f = (E+Wo)/h
 
haruspex said:
Fairly clearly 'proton' is a typo for 'photon'

Another typo: should be '1ev'

I hope you mean f = (E+Wo)/h

Yes!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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