Question

7. There are actually 2 purple lines in hydrogen's atomic spectra, one at 410 nm that is very hard to see and one at 436 nm. What electronic transition do you think is responsible for the line at 410 mm? Explain

Answer 1

Dear Student,

While doing these type of questions the only thing we need to do is For the given length we need to check the probability in different series.From the formulae try for n1=1 from higher states if wavelength we got is greater than the given  one there is a probability to be in that lymann series.In the problem for n=1 wavlength is less than the given one.So we need to move to n=2 for the first one wavelength we got greater than the given value in question.So the electronic transition mostly would have been in n=2 i.e Balmer series.

I have done the question below .

If something is not clear or need further clarification please feel free to comment here without hesitation.

o Given, we have two spectral lines ulonm & 430nm Asked to find electronic transition responsible for yonon. let us assume from nc2 to nel then we have, n = higher ne Lower state state. AE= e = Rre ( 1 2 2 2 2 2 2 2 2 ) е с 919xe FI X5 wavelength. + saling ty xlot macht, Y = + 1.0974 x 10 x 35 x = 1.215x10m - 22 nm which is not true.

we will go for some high values nas to na * = - 0974% 109 m s ) xl.0974 x 107 Jx25 lo97ux lonx24 x = 0.9 ya x 10 x= au.ax loan Ta au.a nmi Volves ase deereeling in we will go for next serie DE2 ns4 609 Tux1st mix(tot) I = 1.0974x10x3 AE 16 3 1009748107 4.8599x10 He 488.99x 15 anom If we nove little bit from here, we can get och we are nearer

ren ,-2 = a( z ) * = -1.097X16" (c ) -OMTRÉ (26) 0974187 36 8x1oan xlor xa 4.102 xion x= ylo. 2x109 x= 410.2 non ? from nesto electronic transition na is responsible for xylonm 2, n=2 refers Balmer series - our line belongs to Balmer series. & v. - Cymann series 223, &, 23 Taschen saring may s.a4 Bracket & gerig