Question
Chemistry Question on Structure of atom
If the ionisation energy of hydrogen atom is 13.6eV, the energy required to excite it from ground state to the next higher state is approximately
A
3.4eV
B
10.2eV
C
17.2eV
D
13.6eV
Answer
3.4eV
Explanation
Solution
For ionization process: Trnasition is from ni=1 to nf=∞, given:IE =13.6eV Transition energy for ionization process is given as: ΔE=E∞−E1=IE as E∞=0 thus E1=−13.6eV First excited state is n=2 as En=n2E1 E2=22−13.6=−3.4eV Transition energy from ground state i.e. n=1 to first excited state i.e. n=2 in Hydrogen atom is given as: ΔE=E2−E1=−3.4−(−13.6)=10.2eV