# Consider a Hydrogen atom in the ground state

1. Nov 25, 2008

### jcvince17

1. The problem statement, all variables and given/known data
Consider a Hydrogen atom in the ground state

Find:

A- The Kinetic Energy, in eV
B - The Potential Energy, in eV
C - The Total Energy, in eV
D - minimum energy required to remove the electron completely from the atom, in eV
E - What wavelength does a photon with the energy calculated in part (D) have

2. Relevant equations

Vn= (1/Eo)(e2/2nh)
Kn = 1/2 mv2 = (1/E02)(me4/8n2h2)
Un = (1/E02)(me4/4n2h2)

mass of H =1.008

3. The attempt at a solution

I have no idea. I know i need to use the first listed equation to solve for v in order to use the second and find the Kinetic Energy. But how do i do the first equation since i do not know the E0

Last edited: Nov 26, 2008
2. Nov 26, 2008

### Redbelly98

Staff Emeritus
If those equations are from your textbook, then they should have a definition for E0 if not the actual value.

Also, something looks wrong. U should have a greater magnitude (and opposite sign) than K, but you have it the other way around.

3. Nov 26, 2008

### jcvince17

I fixed my equations.

now using the first equation to find K. I used the following values:

E0 = 8.854x10-12
m = 1.66*10-27 (i think thats right for Hydrogen)
e = 1.602x10-19
h = 6.626x10-34
n = 1

Kn = (1/E02)(me4/8n2h2)

plug in the values and i get K = 1.28x1022 x 12.17

K = 1.55x1023

but that is wrong. where am i messing up? I only have one try at the answer left.

4. Nov 27, 2008

### jcvince17

what is the correct mass of Hydrogen. I am using 1.66x10^-27. But when i do so and multiply it by e^4 i get 0. which makes everything 0 and this is incorrect.

5. Nov 27, 2008

### Redbelly98

Staff Emeritus
K and U should be a lot smaller than 1 Joule; the answer should have 10 raised to a negative exponent. It may be that you divided where you should be multiplying, or vice-versa, when you put the numbers into your calculator.

I get 1.67 x 10^-27 kg, but that's a small difference and shouldn't really matter.

You could try separating the number from the exponent part, and do those calculations separately. For example:

(6 x 10^8) x (2 x 10^-19)^2 / (4 x 10^-4)
= [ 6 x 2^2 / 4 ] x 10^[8 - (19x2) + 4]
= 6 x 10^-26

That way, the extremely negative exponents won't turn your answer into 0.

Also, it's good practice to include the units in the calculations. This is one way to check that (1) the formulas are consistent and (2) you are multiplying or dividing the terms correctly.

Good luck!

6. Nov 27, 2008

### jcvince17

i tried doing the exponents seperately:
for the me^4 part of equation

(1.67 x 10^-27) x (1.602 x 10^-19)^4
= [ 1.67 x 1.602^4 ] x 10^[-27- (19x4]
= 10.99 x 10^-103

that looks very wrong. and when i try to divide/multiply it by something i get an error in the calc's.

now what am i doing wrong?

another member of another forum suggested another equation:
Kn = 1/n2 * 13.588 eV

if i use this the n will be 1 therefore my Kn is 13.588......?

he also pointed out i am using the wrong mass. i am using mass of atom and i should be using the mass of electron.

Last edited: Nov 27, 2008
7. Nov 27, 2008

### jcvince17

well i am trying to work out the units only to make sure i have the equation right. but i am not ending with eV

E0 = C^2/N x m
m = kg
e = C
h = J x s (or N x m x s)
n = unitless

i get:

(1/m)(kg x C^2 / s)

i cant convert or cancell anything else out. this problem is driving me insane

8. Nov 27, 2008

### Redbelly98

Staff Emeritus
Ah, yes, it should be the electron mass. Sorry I didn't catch that sooner.

If you can just use the 13.6 eV value (or 13.588 eV?) in the formula, that is certainly a lot easier.

I don't have time at the moment to check your units calculation, but note the units of εo are C^2 / (N m^2).

9. Nov 27, 2008

### jcvince17

i found a new equation in the book.

K = 13.6 ev / n^2

U = -27.2 eV / n^2

E = K + U

i knew E = 13.6.

I finally got them all to work out without having to deal with the error's in the exponents.

10. Nov 27, 2008

### jcvince17

now i solved part D and Part E.

thanks

Last edited: Nov 28, 2008