A hydrogen like atom (atomic number `Z`) is in a higher excited state of quantum number `6`. The excited atom can make a transition to the first excited state by successively emitting two photons of energies `10.2eV` and `17.0eV` respectively . Determine the value of `X` . (ionization energy of H-atom is `13.6eV`.)

Answer

C

Explanation

for the first excited stare , `n=2`. Hence
` -13.6eV(1/(6^(2))-1/(2^(2)))z^(2)=10.20+17.00=27.20eV`
` -(1/36-1/4)Z^(2)=(27.20)/(13.6)Rightarrow Z^(2)=9` or `Z=3`