Question

Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelengths λ1:λ2 emitted in the two cases is

Answer 1

《《《 Heyyy 》》》

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Answer 》 The ratio of the wavelengths λ1:λ2 emitted in the two cases is 20 /7..

Full Solution Step by step 》》》》

Here For Wavelength λ1

n1 = 4 & n2 = 3

& for λ2 n1 = 3 & n2 = 2

We have hc/ λ = -13.6 [1/n2^2 - 1/ n1^2 ]

So for λ1

hc / λ1 = -13.6 [1/4^2 - 1/ 3^2 ]

hc / λ1 = -13.6 [7/144] .. . ..........(i)

similarly for λ2

hc / λ1 = -13.6 [1/3^2 - 1/ 2^2 ]

hc / λ1 = -13.6 [5/36] ............(ii)

Hence from equation (i) & (ii).. we get

λ1 /λ2= 20/7 Answer ♥♥♥

Hope helped ♥♥♥

Answer 2

Wavelength observed from transition of ni to nff is 1 [1n2f−1n2i]

For λ\(_1\), n\(_i\) = 4, n\(_f\) = 3

For λ\(_2\), n\ (_i\) = 3, n\(_f\) = 2

λ\(_2\) : λ \(_2\) = 20 : 7

When the electron jumps from the second excited state to the next level then it would cause the new aspects for creating the wavelength that would give more support.