In the normal hydrogen atom the electron is in its lowest energy state, which is

Question

In the normal hydrogen atom the electron is in its lowest energy state, which is called the ground state of the atom. The maximum electronic energy that a hydrogen atom can have is 0 KJ/mole at which point the electron would essentially be removed from the atom and it would become a H+ ion.

How much energy is kilojoules per mole does it take to ionize an H atom? The ionization evergy of hydrogen is often expressed in units other than KJ/mole. What would it be in joules per atom? in electron volts per atom? (1cv=1.602*10-19).

Explanation / Answer

We know that the wavelength of light needed to exite the electron from n=1 to n= infinity can be calculated as

1/ = R [1/n12 - 1/n22]

R = Raydberg's constant = 1.09737×10^7 m-1

on putting n1 = 1 and n2= infinity

1/ = R

So = 1/R = 1 / 1.09737×10^7 = 0.9112 X 10^-7 metres

Energy needed = hc /

h = 6.626×10^-34 J s     c= 3X 10^8 m /s

Energy = 6.626×10^-34 X 3X 10^8 / 0.9112 X 10^-7

Energy = 21.815 X 10^ -19 joules / atom

So for one mole

Energy = 21.815 X 10^ -19 X 6.023 X 10^23 = 131.39 X 10^4 joules / mole = 1313.9 KJ / mole

1 ev = 1.602*10-19 Joules

So 21.815 X 10^ -19 joules = 21.815 X 10^ -19 joules X 1 / 1.602*10-19 Joules ev

So energy in electron volts / atom = 13.617 ev / atom

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