Doubly ionized lithium Li2+ (Z = 3) and triply ionized beryllium Be3+ (Z = 4) each...

Doubly ionized lithium Li2+ (Z = 3) and triply ionized beryllium Be3+ (Z = 4) each emit a line spectrum. For a certain series of lines in the lithium spectrum, the shortest wavelength is 40.5 nm. For the same series of lines in the beryllium spectrum, what is the shortest wavelength?

Expert Solution

The emission (or absorption spectra) of H-like atoms is given by the Rydberg formula as

where R = Rydbergs constant

n_f = principal quantum number of the level to which transition is made.

n_i = principal quantum number of the level from which transition is made.

= wavelength of photon emitted(or absorbed).

For a particular series, n_f is fixed while as n_i changes. Shortest wavelength for a particular series corresponds to n_i = . So, for the shortest wavelength of a series

So

genius_generous answered 2 years ago

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