Algebra & Physics: When Rules Don't Apply

  • Thread starter Thread starter madah12
  • Start date Start date
  • Tags Tags
    Algebra Physics
Click For Summary

Homework Help Overview

The discussion revolves around the application of algebra in the context of the photoelectric effect, specifically questioning when algebraic rules may not hold true in physics. The original poster expresses confusion regarding the relationship between energy, kinetic energy, and the work function in this phenomenon.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of doubling energy in the photoelectric effect and question how the work function, which is a constant, interacts with changes in energy and kinetic energy. There are attempts to clarify the relationships between the variables involved, particularly focusing on whether algebraic manipulations are valid in this context.

Discussion Status

The discussion is ongoing, with participants providing insights and clarifications regarding the relationships between energy, kinetic energy, and the work function. Some participants suggest that the original algebraic reasoning may be flawed, while others attempt to clarify the correct interpretation of the equations involved.

Contextual Notes

There is a noted lack of consensus on the correct application of algebra in this scenario, with participants questioning the assumptions made about the constants and variables in the equations. The original poster acknowledges limited knowledge of the photoelectric effect, which may influence the discussion.

madah12
Messages
326
Reaction score
1

Homework Statement


When does the rules of algebra not apply in physics?
I mean in photo electric effect E = K + [tex]\Phi[/tex] right?
and phi is a constant
so if we have twice the energy 2E = 2K + 2 [tex]\Phi[/tex]
and then E2=K2 + [tex]\Phi[/tex]2
and then [tex]\Phi[/tex]2 = 2E1 - 2K1
[tex]\Phi[/tex]2 /2 =E1-K1 = [tex]\Phi[/tex]2/2 = [tex]\Phi[/tex]1 so phi changed...
but phi is a constant right dependent only on the type of material so it won't change? (I have very little knowledge in photo electric effects this is mainly how and when algebra does not apply in physics)

Homework Equations





The Attempt at a Solution

 
Last edited:
Physics news on Phys.org
Sorry I noticed that I didn't write everything when I made the original post so my argument made no sense I edited it .
 
It's like if you feed one puppy and one kitty on one, well, cake (!?), your puppy and kitty are full; then if you want to throw away 2 cakes in a meaningful way (:biggrin:), then you must get another puppy and another kitty, so 2 puppies and 2 kitties totally for 2 cakes :wink:
If we understand one cake as E, one kitty as K and one puppy as P (let P denote phi, I'm lazy), then actually P2 = 2 puppies. That is, puppy is still puppy, it doesn't grow bigger or smaller, just the number of puppies increases. Likewise, P2 doesn't stand for the work function; P2 is the amount of energy given up by 2 electrons, while P1 (or P) is the amount of energy given up by 1 electron. It's just that we define the work function = the amount of energy given up by 1 electron, or simply P.

EDIT: In conclusion, algebra in particular and math in general concur with physics (or is it the way around, that physics concurs with them? :rolleyes:). I mean, in this case, math does apply.
 
hikaru1221 said:
It's like if you feed one puppy and one kitty on one, well, cake (!?), your puppy and kitty are full; then if you want to throw away 2 cakes in a meaningful way (:biggrin:), then you must get another puppy and another kitty, so 2 puppies and 2 kitties totally for 2 cakes :wink:
If we understand one cake as E, one kitty as K and one puppy as P (let P denote phi, I'm lazy), then actually P2 = 2 puppies. That is, puppy is still puppy, it doesn't grow bigger or smaller, just the number of puppies increases. Likewise, P2 doesn't stand for the work function; P2 is the amount of energy given up by 2 electrons, while P1 (or P) is the amount of energy given up by 1 electron. It's just that we define the work function = the amount of energy given up by 1 electron, or simply P.

EDIT: In conclusion, algebra in particular and math in general concur with physics (or is it the way around, that physics concurs with them? :rolleyes:). I mean, in this case, math does apply.

Umm thanks for the reply. But I shamefully say that I didn't get it completely I mean let say we want the kinetic energy K2 we know where we increased the energy of the photons
we know the values of E , K , phi in the original case and we doubled E
by using another type of light
would K be doubled and phi? I mean we didn't change anything in the material it self but phi can't be doubled because it is constant so the algebriac equation won't hold right?
because unless K2=2K1 and phi2 = 2phi 1 , E2 won't be equal to 2E1
 
Ah, I see. So that really is a problem with math. I mean, your math :frown:
So first you have an electron with kinetic energy K, so you need a photon of energy E. The relation is: E = K + P. Then if you want another electron with 2K, then what you need is NOT a photon of energy 2E! That is, if you shoot a photon of energy 2E, you won't get an electron of 2K.
(1): In order to get an electron of 2K, what you need is a photon of (2K + P), which is NOT equal to 2E.
(2): If you shoot a photon of 2E, what you get is an electron of (2E - P), which is NOT equal to 2K.
And what the mathematical equation 2E = 2K + 2P means, I have already explained. That is the case when you shoot 2 photons of E and consequently, get back 2 electrons of K. This case is not the same as the two cases (1) and (2) above.
 
hikaru1221 said:
Ah, I see. So that really is a problem with math. I mean, your math :frown:
So first you have an electron with kinetic energy K, so you need a photon of energy E. The relation is: E = K + P. Then if you want another electron with 2K, then what you need is NOT a photon of energy 2E! That is, if you shoot a photon of energy 2E, you won't get an electron of 2K.
(1): In order to get an electron of 2K, what you need is a photon of (2K + P), which is NOT equal to 2E.
(2): If you shoot a photon of 2E, what you get is an electron of (2E - P), which is NOT equal to 2K.
And what the mathematical equation 2E = 2K + 2P means, I have already explained. That is the case when you shoot 2 photons of E and consequently, get back 2 electrons of K. This case is not the same as the two cases (1) and (2) above.
Ok I get it now thanks.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
Replies
4
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K