Find distance electron travels before turning around

  • Thread starter Thread starter penguinnnnnx5
  • Start date Start date
  • Tags Tags
    Electron Turning
AI Thread Summary
The discussion revolves around calculating the distance an electron in the hydrogen molecular ion H1,2 travels before turning around, given its initial speed and the fixed position of protons. The participant initially derived an incorrect distance of approximately 0.447 nm, while the correct answer is 3.0 * 10^-3 nm. Key equations involving potential energy (PE) and kinetic energy (KE) were discussed, with emphasis on the correct setup of the energy conservation equation. The importance of using absolute values for charge and proper notation for distances was highlighted to avoid confusion. Ultimately, the correct approach led to resolving the problem and confirming the accurate distance traveled by the electron.
penguinnnnnx5
Messages
36
Reaction score
0

Homework Statement


The hydrogen molecular ion H1,2, with one electron and two protons, is the simplest molecule. The equilibrium spacing between the protons is 0.11 nm. Suppose the electron is at the midpoint between the protons and moving at 1.5 * 10^6 m/s perpendicular to a line between the protons.

How far (in nm) does the electron move before reaching a turning point? Because of their larger mass, the protons remain fixed during this interval of time.

Homework Equations



PE = kQq/r
KE = 1/2 mv2
W = KE + PE
k = 9*109

The Attempt at a Solution



I first drew a diagram to get an idea of how to approach this problem.

Let x2 = .055 nm

R = √(D2 + x2)

-ΔKE = ΔPE
-(-KEi) = PEf

1/2 mv2 = kQq/R + kQq/R = 2kQq/R

R = 4kQq/(mv2) = √(D2 + x2)

D2 + x2 = [4kQq/(mv2)]2

D = √( [4kQq/(mv2)]2 - x2 )

D = ~.447 nm

This answer, however, is wrong. The correct answer is 3.0 * 10-3 nm.

I tried integrating from F, but I still found the same equation for PE. Am I doing something incorrect? Please help as I need to know this by tomorrow for my midterm. Thank you for any help received!
 

Attachments

  • Physics.png
    Physics.png
    2.9 KB · Views: 589
Physics news on Phys.org
So the KE = -PE method is correct? If it is, then I'm not sure what I'm doing wrong at all.

The final answer I have does not match up to the correct answer. It's even off by a factor of ten.
 
PE+KE = const. The potential energy is not zero at the midpoint. Change of potential energy= final KE.

ehild
 
So the set-up should be

PEf - PEi = KEf

2KQq (1/R - 1/x) = 1/2 mv^2 ?
 
What is x2 and why is it 0.055 nm?
 
penguinnnnnx5 said:
So the set-up should be

PEf - PEi = KEf

2KQq (1/R - 1/x) = 1/2 mv^2 ?

yes.

ehild
 
voko said:
What is x2 and why is it 0.055 nm?

x is just a value I used to clean up the work before plugging everything in.

The problem states that the electron travels in the path of a perpendicular bisector of an imaginary line connecting the two protons. If you were to find the radius of that, then it'd be half of the length of the line squared plus the distance traveled squared under a radical. Rather than saying Sqrt( L^2/4 + D^2), it seemed cleaner to say Sqrt( x^2 + D^2) and make x = (1/2)L, L being the length of the line connecting the two protons.
 
ehild said:
yes.

ehild

So I have something like this:

2kQq(1/R - 1/x) = 1/2 mv2

(x-R)/(xR) = (mv2)/(4kQ2) <--- because Q=q=1.6*10-19

(xR)/(x-R) = (4kQ2)/(mv2)

mv2xR = 4kQ2x - 4kQ2R

(mv2x + 4kQ2)R = 4kQ2

R = (4kQ2)/(mv2x+4kQ2)

√(x2+D2) = (4kQ2)/(mv2x+4kQ2)

D2 = (4kQ2)/(mv2x+4kQ2)2 - x2

D = √[(4kQ2)/(mv2x+4kQ2)2 - x2]

However, when I plug in all the values:
x = .055 * 10-9 m
Q = 1.6 * 10-19 C
m = 9.1 * 10-31 kg
v = 1.5 * 106 m/s

I get the square root of a negative number.

Edit:

The set up is supposed to be

KEi = PEi - PEfBUT it should look like this:

-2kQq(1/R-1/x) = 1/2 mv2

2kQq(1/x-1/R) = 1/2 mv2 because Q (the source charges) is 1.6 * 10-19 while q = -1.6 * 10-19

(R-x)/(xR) = (mv2)/(4kQ2)

(xR)/(R-x) = (4kQ2)/(mv2)

mv2xR = 4kQ2R - 4kQ2x

(mv2x - 4kQ2)R = 4kQ2

R = (4kQ2)/(mv2x-4kQ2)

√(x2+D2) = (4kQ2)/(mv2x-4kQ2)

D2 = (4kQ2)/(mv2x-4kQ2)2 - x2

D = √[(4kQ2)/(mv2x-4kQ2)2 - x2]

Plug everything in, and you should be able to get 3.00131072 * 10^-2 nm
 
Last edited:
penguinnnnnx5 said:
x is just a value I used to clean up the work before plugging everything in.

The problem states that the electron travels in the path of a perpendicular bisector of an imaginary line connecting the two protons. If you were to find the radius of that, then it'd be half of the length of the line squared plus the distance traveled squared under a radical. Rather than saying Sqrt( L^2/4 + D^2), it seemed cleaner to say Sqrt( x^2 + D^2) and make x = (1/2)L, L being the length of the line connecting the two protons.

But why is its SQUARE equal to 0.055 nm?
 
  • #10
voko said:
But why is its SQUARE equal to 0.055 nm?

Oh, I'm sorry. That was a typo in the introduction of the problem. It is supposed to be x = .055 nm.

I have not been using the wrong value for x or x2 though, so I am still a little lost.

Edit:

Refer to my previous post as the answer and work is in that post.
 
Last edited:
  • #11
penguinnnnnx5 said:
because Q=q=1.6*10-19

Is that really so?
 
  • #12
penguinnnnnx5 said:
So I have something like this:

2kQq(1/R - 1/x) = 1/2 mv2

It is
2kQq(1/x - 1/R) = 1/2 mv, as KE(initial)=PE(final)-PE(initial). But qQ is a negative number, so your equation is correct if you use the absolute value for the electron charge.

Previously you called the vertical displacement "x". It is better to denote the distance from a proton by "d", for example.

Do not complicate the calculations. Plug in the data, calculate 1/R-1/x. Do not round too much.
(1/R-1/x) can be also written as 1/R(1-R/x).

ehild
 
  • #13
ehild said:
It is
2kQq(1/x - 1/R) = 1/2 mv, as KE(initial)=PE(final)-PE(initial). But qQ is a negative number, so your equation is correct if you use the absolute value for the electron charge.

Previously you called the vertical displacement "x". It is better to denote the distance from a proton by "d", for example.

Do not complicate the calculations. Plug in the data, calculate 1/R-1/x. Do not round too much.
(1/R-1/x) can be also written as 1/R(1-R/x).

ehild

Thank you. As you see above, I found the correct answer. "x" was not set as the vertical displacement, however. Instead, I set "x" as half of the length of the line connecting the two protons and "D" as the distance from the midpoint of the referred line.

I also now see why it is (1/x - 1/R) due to your explanation. As always, you seem to have fully answered my questions without giving away the answers, allowing me to figure out the entire problem. Thank you so much for all the times you have helped me, ehild! :)
 
  • #14
voko said:
Is that really so?

That was my greatest error, as ehild also pointed out. Thank you for your time and effort in helping me understand this problem!
 
  • #15
Well, sorry, I forgot your notations. The PE(final)-PE(initial)=2kqQ(1/R-1/x) , your equation was correct. But q is negative.
Have you got 0.003 nm really? I could not get the same result...

ehild
 
  • #16
ehild said:
Well, sorry, I forgot your notations. The PE(final)-PE(initial)=2kqQ(1/R-1/x) , your equation was correct. But q is negative.
Have you got 0.003 nm really? I could not get the same result...

ehild

Yes I have!
Like you said, since q is negative, the (1/R-1/x) becomes (1/x-1/R) instead.
So,

PEf - PEi = 1/2 mv2

2kQq(1/R-1/x) = 1/2 mv2

-2kQ2(1/R-1/x) = 1/2 mv2 Since q = -Q

2kQ2(1/x-1/R) = 1/2 mv2

1/x - 1/R = (mv2)/(4kQ2)

(R-x)/xR = (mv2)/(4kQ2)

xR/(R-x) = (4kQ2)/(mv2)

(mv2) xR = (4kQ2)R - (4kQ2)x

[(mv2)x - (4kQ2)]R = -(4kQ2)x

R = -(4kQ2)x/[(mv2)x - (4kQ2)]

Since R = √(x2+D2)

√(x2+D2) = -(4kQ2)x/[(mv2)x - (4kQ2)]

x2+D2 = [ -(4kQ2)x/[(mv2)x - (4kQ2)] ]2

D2 = [ -(4kQ2)x/[(mv2)x - (4kQ2)] ]2 - x2

D = √ ( [ -(4kQ2)x/[(mv2)x - (4kQ2)] ]2 - x2 )

With this equation, I found .003 nm. This was after multiplying the result with 109 so convert from meters to nanometers, of course.
 
  • #17
OK, I have found my mistake...


KE=(-2kqQ/x)(1-x/R)
So KE=1.0248˙10-18
-2kgQ/x=8.388˙10-18
1-x/R=0.1222...

ehild
 
Back
Top