Electric potential of hydrogen molecular ion

[smoics]

Homework Statement

The hydrogen molecular ion , with one electron and two protons, is the simplest molecule. The equilibrium spacing between the protons is 0.11nm . Suppose the electron is at the midpoint between the protons and moving at 1.5 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. Note, that an accurate description of H 1/2 requires quantum mechanics. Even so, a classical calculation like this provides some insight into the molecule.

Homework Equations

vf^2=vi^2+2adeltax
F=qE=ma

The Attempt at a Solution

deltax=-mvi^2/2qE=kq/r=E
And q=1.6*10^-19 and mass of proton=.67*10^-27. r=.055*10^-9 m. Is this the correct set-up? When I plug in the numbers, I keep getting weird numbers, but I think my methodology is correct...

 

[tiny-tim]

hi smoics!

(have a delta: ∆ and try using the X² icon just above the Reply box )

deltax=-mvi^2/2qE=kq/r=E
And q=1.6*10^-19 and mass of proton=.67*10^-27. r=.055*10^-9 m. Is this the correct set-up? When I plug in the numbers, I keep getting weird numbers, but I think my methodology is correct...

you seem to have only one value of r

show us your full calculations

 

[smoics]

Okay, are you referencing electric field instead of force? Or using another equation that uses r?

Here's what I'm trying to do:

a=qE/m
a= ((9E9*1.6E-19)/(.055E-9)²/(9.1E-31)
a=5.23E-11

0=1.5E6²+2(5.23E41)$$\Deltax$$
x=-2.2E-21 nm

And I'm sorry there is such a big space before my last line--I tried to edit it, but it doesn't exist when I edit the message (there is no extra space when I edit the message).

 

[tiny-tim]

hi smoics!

(just got up :zzz: …)

Okay, are you referencing electric field instead of force? Or using another equation that uses r?

Here's what I'm trying to do:

a=qE/m
a= ((9E9*1.6E-19)/(.055E-9)²/(9.1E-31)
a=5.23E-11

0=1.5E6²+2(5.23E41)$$\Delta x$$


0=1.5E6²+2(5.23E41)$$\Deltax$$
x=-2.2E-21 nm

And I'm sorry there is such a big space before my last line--I tried to edit it, but it doesn't exist when I edit the message (there is no extra space when I edit the message).

(the LaTeX here has brain failure if you don't leave a space after "\Delta" …

see the code for the above )

sorry, i still don't understand what you're doing

you need a simple conservation of energy equation, KE(r₀) = ∆PE

 

[smoics]

I was trying to solve for the accel. and then use kinematics to find x. I'll try conservation of energy instead:

I find the KE, by using 1/2=mv^2. And then this is equal to the potential energy at the turning point, where KE is 0. U=qV. Solve for V and then add the potential of both protons. Solve for E: E=kq/r^2. E=V/d, solving for d. Is this correct?

 

[tiny-tim]

I'll try conservation of energy instead:

I find the KE, by using 1/2=mv^2. And then this is equal to the potential energy at the turning point, where KE is 0.

yup!

U=qV. Solve for V and then add the potential of both protons. Solve for E: E=kq/r^2. E=V/d, solving for d. Is this correct?

nooo …

(yes, electric field is E= kq/r², but that's not constant, so you can't use V = Er, you have to use V = ∫ Edr)

PE = -kq/r

 

[smoics]

You wrote that PE=-kq/r, but isn't PE at the turning point equal to the original KE?

Isn't there a way to solve it without integrating? Everything we've done in this class has been possible to do without calculus (I got AP credit for the calc I took in high school, so my memory of calculus is minimal).

Can I use W=FXd? I don't think so...

Thanks!

 

[tiny-tim]

You wrote that PE=-kq/r, but isn't PE at the turning point equal to the original KE?

KE + PE = constant …

so 1/2 mv² - kq/r₀ = 0 - kq/r₁

Isn't there a way to solve it without integrating?

yes! …

PE = -kq/r ​

 

[smoics]

1.02E-18-((9E9)(1.6E-19)/(.055E-9))=-(9E9)(1.6E-19)/r
r=-5.5E-2 nm This is incorrect. Am I using the wrong r₀ for the spacing b/w the electron and proton? It should be fairly easy to solve if I use the right numbers...

 

[tiny-tim]

i'm finding it very difficult to understand what all these numbers mean

have you remembered that there are two protons, and that r₁ is the distance from each proton, not the distance from the starting-point?

 

[smoics]

I'm using 0.055 nm as the r0 (half the equilibrium spacing b/w the protons). So the r1 I'm solving for, do I add some length to it to account for the distance from starting point?

I should have scanned in my work, it's easier than typing it...

 

[smoics]

I wasn't accounting for the two protons, by multiplying kq/r by 2. Thanks!