Question

In hydrogen atom the energy difference between
the states n=2 and n= 3 is E eV. The ionisation
potential of H atom is -
(1) 3.2 E
(2) 5.6 E
(3) 7.2 E
(4) 13.2 E

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Answer 1

About #"1.89 eV"# is released for the absorption, or absorbed for the emission process between #n = 3# and #n = 2#. That means it takes about #1.89# #"eV"# of work to accelerate one electron through a one potential difference of one volt.

To be accurate, I don't think that's an ionization potential... that's really the energy used to excite the atom (ionization potential is for removing electron(s) altogether).

I think you're asking about the change in energy going from #n = 3# to #n = 2# in the hydrogen atom? That can be found from the Rydberg equation:

#DeltaE = -"13.6 eV" (1/n_f^2 - 1/n_i^2)#

where:

#"13.6 eV"# is the Rydberg constant in electron volts (or the negative of the hydrogen atom ground state energy).

#n_i# is the initial state's principal quantum number #n#.

#n_f# is the final state's principal quantum number #n#.

So, if you wanted the final state to be #E_3#, then:

#DeltaE = color(blue)(E_3 - E_2) = -"13.6 eV" (1/3^2 - 1/2^2)#

#=# #color(blue)("1.89 eV")#

Answer 2

Answer:7.2E

Explanation:

E= E3 - E2

E= 1.89

I.E. = 13.6 E / 1.89

     = 7.2E