1.

If the energy levels of a hypothetical one electron atom are shown in the figure then find the INCORRECT option. n = oo _____________________________ 0 eV n = 5_____________________________ -0.80 eV n = 4_____________________________ -1.45 eV n=3_____________________________ -3.08 eV n=2_____________________________ -5.30 eV n=1_____________________________ -15.6 eV

Answer»

The ionization potential of this atom is 15.6 V
The short wavelength limit of the series terminating at n = 2 is `2339 Å`
The excitation potential for the STATE n = 2 is 12.52 V
WAVE number of the photon emitted for the transition n = 3 to n = 1 is `1.009xx10^(7)m^(-1)`

Solution :For ionization, electron from n = 1 must be moved to `n = oo`
`THEREFORE` ionization potential `= 15.6 V. lambda_("min")` for series terminating at n = 2 is `(hc)/(5.3)`
For excitation to `n = 2, Delta E = 10.3 eV`
For transition n = 3 to n = 1.
`Delta E = 12.52 eV`.
`therefore lambda = (hc)/(12.52)RARR` wave no. `((1)/(lambda))=(12.52)/(hc)`


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