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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)

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The correct Answer is:
B, C, D
(a) Total 6 lines are emitted. Therefore, `(n(n-1))/(2) = 6 or n = 4 `
So, transition is taking place between mth energy state and (m+3) th energy state.
`E_m = - 0.85 eV or -13.6 ((Z)^2)/(m^2)) =- 0.85`
or `Z/m = 0.25` …(i)
Similarly, `-13.6(Z^2)/((m+3)^2 = -0.544`
or (Z)/((m+3)) = 0.2` ...(ii)
Solving Eqs. (i) and (ii) for Z and m, we get
`m=12and Z =3 `
(b) Smallest wavelength corresponds to maximum difference of energies which is obviously
`E_3 + 3 - E_m` `:. Delta E_max = - 0.544 - (- 0.85) = 0.306 eV`
`:. lambda_(min) = (hc)/(DeltaE_(max) = (1240)/(0.306) = 4052.3mm`
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