# Homework Help: Potential Energy of hydrogen atom in ground state is 0, find energy in 1st exc state

1. Nov 11, 2012

### nishantve1

1. The problem statement, all variables and given/known data
What is the energy of H atom in the first excited state if the potential energy in the ground state is taken to be 0 ?

2. Relevant equations
Usually the energy of H atom in the ground state is -13.6eV
and in the 1st excited state is -10.2eV
E(n) = πme2/8ε2h2

The answer provided is : 23.8eV

3. The attempt at a solution

So I went with how the energy of the electron is derived from ground up
Knowing that
Total energy = potential energy + Kinetic energy
potential energy due to the positive charge on nucleus and negative on electron
so if the potential energy is taken to be 0 then the only energy left is the kinetic one so
the energy would just be
1/2 mvv
then I substituted
v = e2/2∏hn

then I used the formula
ΔE = 13.6(3/4)

but everything is now messed up and I ended up with a gross looking equation , I know something doesn't makes sense above may be nothing does .

P.S : the answer is 23.8 and 13.6 + 10.2 = 23.8
but how can I possibly do this ?

2. Nov 11, 2012

### TSny

Re: Potential Energy of hydrogen atom in ground state is 0, find energy in 1st exc st

Check the value of the energy of the 1st excited state.

3. Nov 11, 2012

### nishantve1

Re: Potential Energy of hydrogen atom in ground state is 0, find energy in 1st exc st

oops! 10.2eV is required to excite the electron from the ground to the first excited state.

4. Nov 11, 2012

### TSny

Re: Potential Energy of hydrogen atom in ground state is 0, find energy in 1st exc st

These don't look correct to me.

5. Nov 11, 2012

### nishantve1

Yeah it isn't
In the E e^2 should be e^4

And in velocity pi should be replaced by epsilon

I posted this from my phone .My bad

6. Nov 11, 2012

### Abhinav R

Re: Potential Energy of hydrogen atom in ground state is 0, find energy in 1st exc st

Ok,to find the potential energy of the hydrogen atom in the first excited state excluding fine structures $\Longrightarrow$

To find the energy levels of the quantum states of the hydrogen atom,we must solve Schrodinger's equation,with equation U = -ke²/r substituted for U in that equation.Solving that equation reveals that the energies of the electron's quantum states are given by,
E$_{n}$ = -m$_{e}$e$^{4}$/8ε$_{o}²$h²n² which leads us to $\Longrightarrow$

-13.6 eV/n$^{2}$ , for n=1,2,3,...,

The value of 13.6 eV is called the Rydberg constant and can be found from the Bohr model.

For n = 2 (first excited state) we get E$_{2}$ = -3.4 eV
It means that the energy required to excite an electron in hydrogen atom to its first excited state,is an energy equal to E$_{2}$ - E$_{1}$ = -3.4 eV-(-13.6)eV
= 10.2 eV

→ E$_{n}$ = -mZ$^{2}$e$^{4}$/8ε$_{0}$$^{2}$n$^{2}$h$^{2}$
→ For n=2,Z=2
→ E = -8.5 × 10$^{19}$ eV

7. Nov 11, 2012

### Abhinav R

Re: Potential Energy of hydrogen atom in ground state is 0, find energy in 1st exc st

Ok,to find the potential energy of the hydrogen atom in the first excited state excluding fine structures $\Longrightarrow$

To find the energy levels of the quantum states of the hydrogen atom,we must solve Schrodinger's equation,with equation U = -ke²/r substituted for U in that equation.Solving that equation reveals that the energies of the electron's quantum states are given by,
E$_{n}$ = -m$_{e}$Z$^{2}$e$^{4}$/8ε$_{o}²$h²n² which leads us to $\Longrightarrow$

-13.6 eV/n$^{2}$ , for n=1,2,3,...,

The value of 13.6 eV is called the Rydberg constant and can be found from the Bohr model.

For n = 2 (first excited state) we get E$_{2}$ = -3.4 eV
It means that the energy required to excite an electron in hydrogen atom to its first excited state,is an energy equal to E$_{2}$ - E$_{1}$ = -3.4 eV-(-13.6)eV
= 10.2 eV

→ E$_{n}$ = -m$_{e}$Z$^{2}$e$^{4}$/8ε$_{0}$$^{2}$n$^{2}$h$^{2}$
→ For n=2,Z=2
→ E = -8.5 eV

I have done this problem including Z therefore my answer is -8.5 eV.Now adding them 10.2 eV + (-8.5)eV
= 1.7 eV