A hydrogen - like atom (described by the Bohr model) is observed to emit six wavelength , originating from all possible transitions between `- 0.85 eV and -0.544 eV` (including both these values )
(a) Find the atomic number of the atom
(b) Calculate the smallest wavelength emitted in these transitions .
(Take `hc = 1240 eV - nm`, ground state energy of hydrogen atom `= 13.6 eV)`

Answer

(i) `Z = 3`, (ii) `4052.3 nm`

Explanation

(i) total 6 lines are emitted.
Therefore `(n(n-1))/(2)=6 implies n=4`
So, transition is taking place between `m^(th)` energy state and `(m+3)^(th)` energy state. `E_(m)= -0.85 eV`
`implies -13.6 (Z^(2)/m^(2))= -0.85" "implies Z/m=0.25` …(i)
Similarly `E_(m+3)= -0.544 eV`
`implies -13.6 z^(2)/((m+3)^(2))= -0.544 implies z/((m+3))=0.2` ...(ii)
Solving equation (i) and (ii) for z and m.
We get `m=12` and `z=3`
(ii) Smallest wavelength corresponds to maximum difference of energies which is obviously `E_(m+3)-E_(m)`.
`:. DeltaE_("max")= -0.544-(-0.85)=0.306 eV`
`:. lambda_("min")=(hc)/(DeltaE_("max"))=1240/0.306=4052.3 nm`.