# Atoms

• Ans is 0

• total energy in 1st excited state = +23.8 eV
energy releases when it goes back to ground state = 10.2 eV (it is the amount of energy required to make the transition from n=1 to n=2)
total energy in ground state = +23.8-10.2 = 13.6 eV
kinetic energy in ground state = -13.6 eV
so, potential energy in ground state = 13.6-13.6 = 0

• @vinay-jaiswal if total is 13.6, ke is -13.6 then shudnt potentia be 27.2

• sorry, kinetic energy is always +ve . total energy in ground state = 13.6 eV
total energy = P.E. + K.E.
13.6 = P.E. + 13.6
p.e. = 0

• @vinay-jaiswal but kinetic energy and total energy have opposite signs as written in my textbook

• in the above problem , it is assumed that the total energy is 23.8 eV but in reality it is not the case.

• @vinay-jaiswal this is confusing me a lot please provide elaborate explanation.

• @pranay-singh first, look at the basic concepts of p.e. and k.e.
the energy required to bring an electron from infinity to ground state is potential energy in ground state. it is -ve for attractive force. It is assumed that the p.e. is 0 at infinity. so the p.e. in ground state = -27.2 eV as per the above mentioned assumption.
k.e. in ground state = +13.6 eV
so total energy = -13.6 eV

Now, this question
it is not taking the above mentioned assumption that the p.e. is 0 at infinity.
total energy in 1st excited state = 23.8 eV
energy required for transition from n=1 to n=2 = 10.2 eV
total energy in ground state = 23.8-10.2 = 13.6 eV
k.e. in ground state = 13.6 eV
so p.e. in ground state = t.e. - k.e. = 0

k.e. in ground state will always same as velocity will always be same.
energy required for transition from any state to other is fixed for each level.
p.e. depends on the assumption that at which point we are taking the 0 level p.e.