A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. The excited atom can make a transition to the first excited state by successively emitting two photons of energy 10.2 eV and 17.0 eV, respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV, respectively Determine the values of n and Z. (lonization energy of H-atom = 13.6 eV)

Options

A. 6 and 6
B. 3 and 3
C. 6 and 3
D. 3 and 6

Answer

C

Explanation

In first case , transition is from `n^(th)` state to `2^(nd) (1 ^(st excite)` state.
`:. (10.2 + 17.0) eV = 13.6 xx Z^(2) [(1)/(2^(2))/(1)/n^(2)]`
In `2^(nd)` case , transition is from `n^(th)` state to `3^(rd)`state.
`:. (4.25 + 5.95) eV = 13.6 xx Z^(2) [(1)/(3^(2))/(1)/(n^(2))]`
Solving above equation , we get `n = 6 and Z = 3`.