Energy of H Atom in 1st Excited State - 23.8 eV

[nishantve1]

Homework Statement

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 ?

Homework 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
Bohr's radius = 0.529 angstrom

The answer provided is : 23.8eV

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 mv^v
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 .

Thanks in advance :

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

 

[TSny]

Usually the energy of H atom in the ground state is -13.6eV
and in the 1st excited state is -10.2eV

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

 

[nishantve1]

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

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

 

[TSny]

E(n) = πme²/8ε²h²

v = e²/2∏hn

These don't look correct to me.

 

[nishantve1]

These don't look correct to me.

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

 

[Abhinav R]

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 Schrödinger'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

 

[Abhinav R]

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 Schrödinger'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

But according to your answer,it looks different!