The Bohr equation for the hydrogen atom is En= -2.179 x 10-18J/n2. 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.
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 RH or 2.179 E-18 J/atom
That 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? - E1 and E? = 0 thus you end up with 0 - (-2.179E-18) = 2.179E-18 J/atom
Now 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? - E2 = 0 - (-5.45E-19) = 5.45E-19 J/atom
jumping 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