For a certain hypothetical one electron atom, the wavelength `(in Å)` for the spectral lines for transitions originating at n=p and terminating at n=1 are given by `lambda = (1500 p^2)/(p^2 - 1), where p = 2,3,4`
(a)Find the wavelength of the least energetic and the most energetic photons in this series.
(b) Construct an energy level diagram for this element showing the energies of the lowest three levels.
(c ) What is the ionization potential of this element?

Options

A. `1800 Å`
B. `1500 Å`
C. `1300 Å`
D. `1650 Å`

Answer

B

Explanation

As we know energy of a photon is given by `E = (hc)/(lambda)`
Fron the given condition, `lambda = (1500 p^(2))/(p^(2) - 1)`
Hence, `E = (hc)/(1500) (1 - (1)/(p^(2))) xx 10^(10) J`
`= (hc)/((1500) (1.6 xx 10^(-19))) J (1 - (1)/(p^(2))) xx 10^(10) eV`
`= 8.28 (1 - (1)/(p^(2))) eV`
Hence, energy of nth state is given by
`E_(n) = - (8.28)/(n^(2)) eV`
Maximum energy is released for transition from `p = oo to p = 1`: hence wavelenght of most energitic photon is `1500 Å`.