One of the emission spectral lines for be3 + has a wavelength of 466.4 nm for an electronic transmission that begins in the state with n = 6. what is the principal quantum number of the lower-energy state corresponding to this emission?

Question

Justice Curtis

Physics

27 February, 14:36

One of the emission spectral lines for be3 + has a wavelength of 466.4 nm for an electronic transmission that begins in the state with n = 6. what is the principal quantum number of the lower-energy state corresponding to this emission?

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Answers (1)

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Baldie · Answer 1

The answer is - 6.05 x 10 ^-20. In order to get this answer, you have to use the Rydberg formula of - R (1/n ^2).

Rydberg constant is - 2.178x10 ^ - 18

n = 6

so the solution is

=-2.178 x10 ^ - 18 (1 / 6^2)

=-6.05 x 10 ^ - 20