Hi! Not sure if anyone asked this yet, but for number 1B.27: A bowling ball of mass 8.00kg is rolled down a bowling alley lane at 5.00 +- 5.0 m/s. What is the minimum uncertainty in its position?
I used the normal Heisenberg indeterminacy equation and for delta v I plugged in 10. m/s because I doubled the 5.0. However, the answer key seems to have used the 5.0 instead of 10., so I was wondering if this was a mistake in the book or on my part?
Thank you!
1B. 27 [ENDORSED]
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Re: 1B. 27 [ENDORSED]
This question should be under the solution manual errors on Prof. Lavelle's website. See here: https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14A/Solution_Manual_Errors_7Ed.pdf
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Re: 1B. 27
Hi, can someone explain why the delta v would be 10 instead of 5? If the plus or minus range on the velocity is 5 m/s, wouldn't the delta v just be 5?
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Re: 1B. 27
Hey Sarah!
Delta v would be 10 because there is an uncertainty of 5 on the high end and the low end. For example, if your velocity is 15 m/s, taking the uncertainty into account, it could land anywhere in between 20 m/s as an upper bound and 10 m/s as a lower bound. The difference between the maximum and minimum would be 10 m/s, therefore delta v would be 10!
Hope this makes a bit of sense :)
Delta v would be 10 because there is an uncertainty of 5 on the high end and the low end. For example, if your velocity is 15 m/s, taking the uncertainty into account, it could land anywhere in between 20 m/s as an upper bound and 10 m/s as a lower bound. The difference between the maximum and minimum would be 10 m/s, therefore delta v would be 10!
Hope this makes a bit of sense :)
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