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

• stickplot
In summary: The TI84 calculator is not accurate when it comes to scientific notation. You should be able to enter 5.12x10^-28 without any problems, but it will give you the incorrect answer of 2247866.
stickplot

## 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..

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?

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?
and i also know that r is the radius
and v is what were looking for

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​

Good luck

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

The electron mass is wrong.

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?

Yes. kg, not grams, are the base mass unit in the SI metric system.

Example: 1 Newton = 1 kg·m/s²

Last edited:
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.

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.

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:
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.

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?

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"

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

## 1. What is the atomic model?

The atomic model is a scientific representation of the structure and behavior of an atom. It helps us understand the fundamental nature of matter and how atoms interact with each other.

## 2. Who proposed the first atomic model?

The first atomic model was proposed by the Greek philosopher Democritus in the 5th century BC. He believed that all matter was made up of tiny, indivisible particles called atoms.

## 3. What is Coloumb's Law?

Coloumb's Law is a fundamental law in physics that describes the force of attraction or repulsion between two charged particles. It states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

## 4. How does Coloumb's Law relate to atomic models?

Coloumb's Law is an essential part of the atomic model as it explains the forces that hold the atom together. It helps us understand the interactions between charged particles, such as protons and electrons, within the atom.

## 5. What are the limitations of the atomic model and Coloumb's Law?

One of the main limitations of the atomic model is that it is a simplified representation of the complex nature of atoms. It does not account for the wave-like behavior of particles in the atom, which is described by quantum mechanics. Coloumb's Law also has limitations, as it does not take into account the effects of special relativity at high speeds or the behavior of particles at the subatomic level.

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