Calculating uncertainty for hydrogen atom

[crybllrd]

Homework Statement

For a object of mass m, Heisenberg’s uncertainty principle relates the uncertainty in the object’s position Δx to the uncertainty in the object’s speed Δv:
(Δx)(Δv) ≥ (h divided by (4)(pi)(m))
where h is Planck’s constant.
Calculate the minimum uncertainty in the speed of a tennis ball of mass 0.058 kg, assuming that the uncertainty in its position is approximately equal to its own diameter of 6.5 cm. If you assume the tennis ball has a speed equal to the uncertainty value you calculated, how long would it take for the ball to travel a distance equal to its own size? Based on this, do you feel we can ever say where a tennis ball is with a reasonable uncertainty?
Repeat all of the above analysis for a hydrogen atom of mass 1.67 × 10−27 kg with diameter 1.06 angstrom.

Homework Equations

h = 6.62606896× 10e-34 J·s

The Attempt at a Solution

First I plug in the numbers to figure out velocity v:
(6.5cm)(Δv) ≥ (6.62606896× 10e-34 J·s divided by (4)(pi)(0.058kg))

(6.5cm)(Δv) ≥ (about) 9

Using basic algebra:

(Δv) ≥ 9/6.5

(Δv) ≥ 1.4

So then I plug 1.4 into (Δv)? What do I solve for? I'm not sure where to go from here.

 

[willem2]

You already have the answer. You forgot about the $10^{-34}$ in your calculation however. Make sure you give the right units.

 

[crybllrd]

Hmmm I don't understand.
What do you mean by I forgot about the 10$$^{-34}$$ in your calculation? Do you mean I need to use that unit (seconds)?

If so, then I think the complete answer is that "it would take 1.4 seconds for the ball to travel a distance equal to its own size."

 

[willem2]

Hmmm I don't understand.
What do you mean by I forgot about the 10$$^{-34}$$ in your calculation? Do you mean I need to use that unit (seconds)?

you used h = 6.62606896 Js, instead of h=6.62606896× 1e-34 Js

If so, then I think the complete answer is that "it would take 1.4 seconds for the ball to travel a distance equal to its own size."

The first part of the question was: Calculate the minimum uncertainty in the speed of a tennis ball That is what you did (except for a factor of 1e-34)

the answer should be a speed, so the units should be in m/s. Convert the size of the ball to meters also.

 

[crybllrd]

OK I converted to meters(6.5cm=.065m). Now I have:

(.065m)(Δv) ≥ (6.62606896× 10e-34 J•s divided by (4)(pi)(0.058kg))

(.065m)(Δv) ≥ (about) 9Js

Using basic algebra:

(Δv) ≥ 9J•s /.065m

(Δv) ≥ 138.46m/s

The minimum uncertainty for the speed of a tennis ball is 138.46m/s

Now I must figure out how long it will take for the ball to travel a distance of its own size at a velocity of 138.46m/s.

Can I just divide .065m by 138.46m/s?
If so, then the ball will travel 4.6 × 10^-4m

 

[willem2]

you still use h = 6.62606896 Js, instead of h=6.62606896× 1e-34 Js

why do you think

(.065m)(Δv) ≥ (6.62606896× 10e-34 J•s divided by (4)(pi)(0.058kg))

implies

(.065m)(Δv) ≥ (about) 9Js

do you really think the minimum velocity uncertainty of tennis balls is 138 m/s ?

 

[crybllrd]

Ahh, ok, I was dividing with the correct number, but I didn't see the "e-4" at the end for a reason that is more suitable for a TI-84 technical forum. :)
Thanks for catching that!
0.0009

(.065m)(Δv) ≥ .0009Js
(Δv) ≥ 0.014m/s

divide .065m by 0.014m/s:
4.64s