Angular velocity of an electron

[nded07]

SOLVED: Angular velocity of an electron

1. The Bohr model of the hydrogen atom pictures the electron as a tiny particle moving in a circular orbit about a stationary proton. In the n = 2 orbit the distance from the proton to the electron is 21.16*10^-11 m, and the linear speed of the electron is 1.09*10^6 m/s.



2. What is the angular velocity of the electron?



3. To find the angular acceleration (ω), I started with v=rω (linear velocity=radius*angular velocity) and arranged it to solve for ω (ω=v/r). Plugging the given values in, I came up with ω=1.09*10^6/21.16*10^-11, and I got 2.43*10^14 (and 5.15*10^15 as well, for some reason) for an answer, which is wrong according to WebAssign. I'm down to my last submission and would really appreciate it if someone could point out what I'm doing wrong.

 

[nded07]

Ah, never mind. Turns out WebAssign wanted the unaltered string of numbers, which runs up to about 5151228733459357.28 rad/s. Thank goodness for Windows basic calculator.

 

[KMjuniormint5]

I don't know how well you know webassign but when plugging in stuff like that say 5.15*10^15 for instance . . . try plugging in 5.15 E 15 (that is the way they want not 5.15*10^15 )

 

[nded07]

I tried that, but it didn't work. Apparently it was looking for as precise a value as possible, and 5150000000000000 wasn't going to cut it.

 

[KMjuniormint5]

that is really weird. . it shouldn't except that large of a number. . but you got the correct answer?

 

[nded07]

Yes, the last answer I submitted registered as being correct. However, looking at the answer key tells me that 5.15e+15 would also have been an acceptable format for the answer. I'm new to the whole WebAssign thing though, and the fact that the question used a different notation from what was expected for the answer certainly threw me off the loop. Live and learn, I suppose; at least I'll know what to expect next time something like this pops up.

On the bright side, the rest of the assignment went fairly smoothly and I managed to get good grade.

 

[KMjuniormint5]

thats sweet . . sounds like you are taking about the same level of physics as I, if you need any help just let me know