The Bohr equation for the hydrogen atom is E_{n}= -2.179 x 10^{-18}J/n^{2}. What is the ionization energy, in J, for the loss of an electron from the lowest excited state?

The answer is 5.45 x 10^{-19}, but I have no idea how to find that.

thanks in advance

lowest excited state is n= 1to n=2

En= -2.179 x 10-18J/n2.

just calculate

E1= -2.179 x 10-18J/1X1

E2- E1 and that will give you energy required to excite the elctron from lowest to 2.

E2= -2.179 x 10-18J/2X2

The ionization energy for hydrogen from the ground state is R

_{H}or 2.179 E-18 J/atomThat is the value that you will find in any table, or if you want it in kj/mol, then it is 1312 kj/mol

The ionization energy is calculated as the difference of the energy levels from n=1 to n=infinity. You excite the electron so high that it does not return to the ground state, it is lost, thus you end up with a +1 ion.

Thus to ionize Hydrogen from the ground state you have to calculate the energy difference from

E

_{?}- E_{1}and E_{?}= 0 thus you end up with 0 - (-2.179E-18) = 2.179E-18 J/atomNow your question is asking for the ionization energy from the first exited state.

The first excited state for hydrogen is n = 2

You need to calculate E

_{?}- E_{2}= 0 - (-5.45E-19) = 5.45E-19 J/atomjumping from n=1 to n=2 will not ionize hydrogen, it will only exite it.

oops my bad

I had shown method for the lowest energy transitions ---n=1 to n=2

in the question it is ionization energy from lowest to the infinity

n= 1 to n= infinity

I