Consider a hypothetical hydrogen like atom. The wavelength in Å for the spectral lines for transitions from n = p to n = 1 are given by λ = 
, where p = 2, 3, 4, …..
Text Solution
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Sol. As we know energy of a photon is given by
E = 
From the given condition λ = 
Hence, E = 
= 
× 10 10 J
= 
× 10 10 eV
= 8.28 
eV
Hence energy of nth state is given by E n = – 
eV
Maximum energy is released for transition from p = ∞ to p = 1; hence wavelength of most energetic photon is 1500 Å.
Least energy is released for transition from n = 2 to n = 1 transition. For p = 2, λ = 2000 Å
The energy level diagram is shown in Fig.
Fig.
The ionization potential corresponds to energy
required to liberate an electron from its ground state,
i.e., ionization energy = 8.28 eV
Hence, ionization potential = 8.28 eV
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