3. Find out the initial energy level (n) of an electron that results in the emi ssion of light of
wavelength 486 nm in the Balmer series?

4

Explanation:

The Balmer series refers to a series of spectral emission lines of the hydrogen atom arising from electronic transitions from any higher level and terminating at the the energy level n= 2.

Using the relation;

1/λ = 1.09737*10^7 (1/2^2 -1/n^2initial)

We now have;

1/486 *10^-9 = 1.09737*10^7 (1/2^2 -1/n^2initial)

0.1875 = 1/2^2 -1/n^2initial

1/n^2initial = 1/2^2 - 0.1875

1/n^2initial = 0.0625

n^2initial = 16

ninitial = 4

the answer is barium and copper bromide