The ionization enthalpy of hydrogen atom is 1.312

Asked by admin @ 31/08/2021 in Science viewed by 346 People

The ionization energy of hydrogen atom is 1.312 into 10 raise to power 6 joule/mol

calculate the energy required to excit and electron in a hydrogen from ground state to its first state

Answered by admin @ 31/08/2021

Answer:

The different energy levels of a Hydrogen atom are given by the equation:

E = - E0/n2

where E0 = -1.312 KJ/mol and n = 1,2,3… and so on.

For ground state n = 1

E1=E0/(1)2

So that the ground state has energy E1= -1.312 KJ/mol

For first excited state n = 2

So E2=-1.312 KJ/mol /4= -0.328 KJ/mole

Energy difference between two states:

ΔE= E2-E1= (-0.328)-(-1.312)

= 0.984 KJ/mol

0.984 KJ/mol energy is required to excite an electron in hydrogen atom from ground to excited state.

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