A hydrogen like atom (described by the Borh model) is observed to emit six wavelength, originating from all possible transitions between a group of levels. These levels have energies 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)

Explanation

Number of lines in emission spectrum from `n=n_(1) "to" n=n_(2) (n_(1) gt n_(2))`
`((n_(1)-n_(2))(n_(1)-n_(2)+1))/(2)=6`
Let `n_(1)-n_(2)=x`
`(x(x+1))/(2)=6 implies x^(2)+x-12=0`
`(x+4)(x-3)=0`
`x=3,x=-4 ("not possible") `
`n_(1)-n_(2)=3 implies n_(1)=n_(2)+3`
Let `n_(2)=n, n_(1)=n+3`
`E_(n)=-13.6(Z^(2))/(n^(2)) eV =-0.85 eV`
`E_(n+3)=-13.6 (Z^(2))/((n+3)^(2)) eV =-0.544 eV`
`(i)//(ii)`
`((n+3)/(n))^(2)=(0.85)/(0.544)=1.5625`
`(n+3)/(n)=1.25=(5)/(4)`
`4n+12=5n implies n=12`
`13.6 (Z^(2))/(n^(2))=0.85 implies (13.6)/((12)^(2)) Z^(2)=0.85`
Z=3
(b) Smallest wavelength will be in the transition
`(n+3) "to" n i.e. 15 "to " 12`
`(1)/(lambda) = Z^(2)R((1)/(12^(2))-(1)/(15^(2)))=(3)^(2)R [(225-144=81)/(144xx225)]`
`lambda=(44.4)/(R ) =(44.4)/(1.097xx10^(7))=40.51xx10^(-7)m =4051 nm`