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Question

Chemistry Question on Structure of atom

If the ionisation energy of hydrogen atom is 13.6eV13.6\,eV, the energy required to excite it from ground state to the next higher state is approximately

A

3.4eV3.4\,eV

B

10.2eV10.2\,eV

C

17.2eV17.2\,eV

D

13.6eV13.6\,eV

Answer

3.4eV3.4\,eV

Explanation

Solution

For ionization process: Trnasition is from ni=1n _{ i }=1 to nf=n _{ f }=\infty, given:IE =13.6eV=13.6 eV Transition energy for ionization process is given as: ΔE=EE1=IE\Delta E = E _{\infty}- E _{1}= IE as E=0E _{\infty}=0 thus E1=13.6eVE _{1}=-13.6 eV First excited state is n=2n =2 as En=E1n2E _{ n }=\frac{ E _{1}}{ n ^{2}} E2=13.622=3.4eVE _{2}=\frac{-13.6}{2^{2}}=-3.4 eV Transition energy from ground state i.e. n=1n =1 to first excited state i.e. n=2n =2 in Hydrogen atom is given as: ΔE=E2E1=3.4(13.6)=10.2eV\Delta E = E _{2}- E _{1}=-3.4-(-13.6)=10.2 eV