(adsbygoogle = window.adsbygoogle || []).push({}); 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) = πme^{2}/8ε^{2}h^{2}

Bohr's radius = 0.529 angstrom

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 mv^{v}

then I substituted

v = e^{2}/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 ?

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

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