What is the Orbital Velocity of an Electron in a Hydrogen Atom?

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Homework Help Overview

The discussion revolves around calculating the orbital velocity of an electron in a hydrogen atom using the planetary model. The original poster provides the radius of the electron's orbit and attempts to apply relevant equations involving electrostatic forces and the mass of the electron.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the use of the equation mv² = k(e²/r) to find the orbital velocity. There are attempts to clarify the values for the mass of the electron and Coulomb's constant, with some questioning the correctness of the initial setup and calculations.

Discussion Status

Participants are actively engaging with the problem, providing guidance on the values needed for calculations and addressing misunderstandings about units and constants. There is a recognition of arithmetic errors, and some participants express concern about the original poster's approach while others offer support and resources for further understanding.

Contextual Notes

There are mentions of confusion regarding the mass of the electron and the proper use of scientific notation in calculations. The original poster notes that their course is online and lacks a textbook reference for certain constants.

stickplot
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Homework Statement



The radius of the orbit of an electron in a hydrogen atom is 0.05 nm. Using the planetary model, calculate the orbital velocity needed by the electron to avoid being pulled into the nucleus by electrostatic attraction. Round up your answer to the nearest tenth.

Homework Equations



mv^2=k(e^2/r)
mass of electron -13.6?? (not sure)
v=velocity
k= coloumb constant
e=1.6x10^-19

The Attempt at a Solution



13.6(v^2)=k(q1q2/r2)(1.6x10^-19/5x10^11 m)
i think the equation is wrong :( I am not sure about coloumbs constant and what it is..
 
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stickplot said:

Homework Equations



mv^2=k(e^2/r)

That much is right.

mass of electron -13.6?? (not sure)

It's OK if you aren't sure of the mass of an electron, but for Jeebus' sake son use your brain. When was the last time you put something on a scale and found that its weight is negative? :-p

You should look up the electron mass. That's not too much to ask.

v=velocity
k= coloumb constant
e=1.6x10^-19

OK

The Attempt at a Solution



13.6(v^2)=k(q1q2/r2)(1.6x10^-19/5x10^11 m)
i think the equation is wrong :( I am not sure about coloumbs constant and what it is..

While you're looking up the electron mass, look up that constant as well, please. They are both in your book.
 
ok. now i understand that its just asking for coloumbs constant and not for the law. but my course is online not textbook, and has never showed me 8.99x10^9Nm2/C2 as coloumbs constant, all it says in the tutorial is that coloumbs constant (k) makes sure the units work out properly.
i understand now that e= charge of electron which is 1.6x10^-19 coloumbs
and i understand now that the mass of an electron is 9.10938188 × 10-31 kilograms but is that what its asking for in the equation? of mv^2=k e^2/r^2?
thanks for your replies
and i also know that r is the radius
and v is what were looking for
thanks for your replies
 
i understand now that the mass of an electron is 9.10938188 × 10-31 kilograms but is that what its asking for in the equation? of mv^2=k e^2/r^2?
The electron mass is the m in that equation. They're not asking for m, but you need it in order to calculate what v is.

EDIT:
p.s. typo? The equation really is
m v^2 / r = k e^2 / r^2​
or
m v^2 = k e^2 / r​
as you had originally.

Good luck :smile:
 
o ok. and btw it was a typo
m v^2 = k e^2 / r
ok so i got everything down now I am just still confused about k do i insert 8.99x10^9Nm2/C2 for k or like i said before all k does is it "makes sure the units work out properly."
 
stickplot said:
... do i insert 8.99x10^9Nm2/C2 for k ...

Yes.
 
how does this look?
m v^2 = k e^2 / r
9.10938188×10^-28(v^2)=8.99x10^9(1.6x10^-19^2/5x10^-11)
9.10938188×10^-28(v^2)=8.99x10^9(5.12x10^-28)
9.10938188×10^-28(v^2)=4.60288x10^-18
v^2=5052900472
v= 71080 m/s
 
  • #10
The electron mass is wrong.
 
  • #11
i put the electron mass in grams i thought it was supposed to be in the basic unit.
so is it supossed to be in kilo? 9.10938188 × 10-31?
 
  • #12
Yes. kg, not grams, are the base mass unit in the SI metric system.

Example: 1 Newton = 1 kg·m/s²
 
Last edited:
  • #13
ok. so that made a big difference in the answer
m v^2 = k e^2 / r
9.10938188×10^-31(v^2)=8.99x10^9(1.6x10^-19^2/5x10^-11)
9.10938188×10^-31(v^2)=8.99x10^9(5.12x10^-28)
9.10938188x10^-31(v^2)=4.60288x10^-18
v^2=505290047x10^-27
v= 7.10837567x10^-14 m/s

how is that?
and its meters per second right.
 
  • #14
I will be honest. It is close to the point where I wonder if you are really trying to solve this problem, or are deliberately making mistakes so that other people (i.e. me) waste their time responding.
 
  • #15
ok thanks i appreciate your honesty.
if you don't want to help its fine but could you at least give me a website that explains because I've been looking and i can't find anything that helps me.
 
Last edited:
  • #16
If I'm wrong about my suspicions, I sincerely apologize.

You have been so close to solving this, but made an arithmetic error somewhere in post #13. Your formula is correct. Your input values (k, m, e, r) are correct. It's just a matter of being careful with the arithmetic at this point.
 
  • #17
o wow. i can see now why you thought i was kidding, i think i got it now.
m v^2 = k e^2 / r
9.10938188×10^-31(v^2)=8.99x10^9(1.6x10^-19^2/5x10^-11)
9.10938188×10^-31(v^2)=8.99x10^9(5.12x10^-28)
9.10938188x10^-31(v^2)=4.60288x10^-18
v^2=5.052900472x10^12
v= 2247866 m/s
and btw i have a ti 84 calc and for some reason whenever i enter scientific notations like 5.12x10^-28 and divide it by something or multiply, it will always give me a different answer and when i do it manually and i insert all the 0s it will give me the right answer, i don't know if its the calc, or something else that I am doing wrong but its a hassle because i always have to insert so many 0s! any advice?
 
  • #18
stickplot said:
v= 2247866 m/s
Yes! And I'm sorry about the misunderstanding.

... and btw i have a ti 84 calc and for some reason whenever i enter scientific notations like 5.12x10^-28 and divide it by something or multiply, it will always give me a different answer and when i do it manually and i insert all the 0s it will give me the right answer, i don't know if its the calc, or something else that I am doing wrong but its a hassle because i always have to insert so many 0s! any advice?

I have a ti 38+, it's probably the same as the ti 84.

To enter 9.1x10^-31, the keystrokes are:

9
.
1
"2nd"
"EE" (NOTE: same key as ",", just about "7" key)
"(-)" (NOTE: not the "-" subtraction key, the "(-)" key is below the "3" key)
3
1

Is that how you are doing it? By the way the "EE" stands for "Enter Exponent", it means "times 10 raised to the exponent to be entered next"
 
  • #19
o ok i think that should work i was doing it by inserting
9
x
10
^
-
31
so i think your way should work.
thanks for the replies and for helping me out with my question
 

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