Calculate the wavelength, in nm, of the light emitted when a Be^3+ ion makes a transition from the n=5 to the n=2 state.

E = (4.98 eV) [1/n_{f}^{2} - 1/n_{i}^{2}]

E=4.98(1/4 -1/25 )

E=1.045 ev

wavelength=E=hc/wavelength

wavelength=1.1865e+3 nm

or

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Did you have a typo in the 5th line? is it suppose to be wavelength= 1.18665 Be^3+ nm ?

no it is calculation by using formula E=hc/wavelength

h=6.626 x 10^{-34} J s c=3.0 x 10^{8}m/s (the speed of light in a vacuum) E=1.045 ev

i put this in ans mistakenly : wavelength=E=hc/wavelength

ok you get it?

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