# Prove the blue light of wavelength, l = 444 nm, is not capable of exciting electrons in Li2+(g) from n = 2 to n = 4?

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### 1 Answer

You need to use the equation that helps you to evaluate the minimum energy of a photon that produces excitation,such that:

`E= KZ^2(1/(n_2)^2 - 1/(n_1)^2)`

`n_2` represents the final state

`n_1` represents the initial state

You also may evaluate the energy that excites the elctrons in `Li^(2+)` such that:

`E = (hc)/lambda`

Equating the expressions above, yields:

`KZ^2(1/(n_2)^2 - 1/(n_1)^2) = (hc)/lambda`

`lambda = (hc)/(KZ^2(1/(n_2)^2 - 1/(n_1)^2))`

`lambda = (6.6*10^(-34))/(2.18*10^(-18)*9*(1/2^2 - 1/4^2))`

`lambda = 53.8 nm`

**Hence, since `lambda = 53.8 nm < 444nm` , the wavelength of blue light, hence, the energy of blue light is not sufficiently to produce the excitation of electrons of `Li^(2+).` **