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 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}²h²n² 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 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}²h²n² 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!