Find distance electron travels before turning around

In summary, the problem involves finding the distance an electron moves before reaching a turning point while traveling between two fixed protons at a certain equilibrium spacing. The equation used is the change in kinetic energy equals the change in potential energy, with the potential energy being calculated using Coulomb's Law. The final answer is 3.00131072 * 10^-2 nm.
  • #1
penguinnnnnx5
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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!
 

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  • #2
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.
 
  • #3
PE+KE = const. The potential energy is not zero at the midpoint. Change of potential energy= final KE.

ehild
 
  • #4
So the set-up should be

PEf - PEi = KEf

2KQq (1/R - 1/x) = 1/2 mv^2 ?
 
  • #5
What is x2 and why is it 0.055 nm?
 
  • #6
penguinnnnnx5 said:
So the set-up should be

PEf - PEi = KEf

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

yes.

ehild
 
  • #7
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.
 
  • #8
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:
  • #9
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
 

Related to Find distance electron travels before turning around

What is the meaning of "Find distance electron travels before turning around"?

The question asks for the distance an electron can travel before changing direction. This is often referred to as the electron's turning point or turnaround point.

What factors affect the distance an electron can travel before turning around?

The distance an electron can travel before turning around is affected by the strength of the electric field it is in, as well as its initial velocity and the mass of the electron.

Is there a formula for calculating the distance an electron travels before turning around?

Yes, the formula is d = (mv2) / qE, where d is the distance, m is the mass of the electron, v is its initial velocity, q is the charge of the electron, and E is the strength of the electric field.

Can the distance an electron travels before turning around be greater than its initial distance?

Yes, it is possible for the electron to travel a greater distance before turning around if the electric field is strong enough to accelerate the electron to a higher velocity.

Why is it important to understand the distance an electron travels before turning around?

Understanding the distance an electron can travel before turning around is important in many areas of science, such as electronics, particle physics, and astronomy. It can also help us understand the behavior of charged particles in electric fields and their interactions with other particles.

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