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# In Hydrogen atom electron transition takes place between L shell and K shell then wavelength of light emitted is

9 years ago

Dear moses christoper,

If an electronic transition takes place from the L Shell to the K Shell i.e., n=2(L Shell) to n=1(K Shell) then the wavelength of the light emitted is given by the Rydberg's formula $\frac{1}{\lambda_{vac}} = R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Where $\lambda_{vac} \!$ is the wavelength of electromagnetic radiation emitted in vacuum, $R\!$ is the Rydberg constant, approximately 1.097 * 107 m-1, $n_1\!$ and $n_2\!$ are integers such that $n_1 < n_2\!$.

Therefore the wavelength according to your problem will be 91.13 nm

Hope this helped you immensely...

Regards,

Godfrey Classic Prince

Please approve my answer if you liked it by clicking on "Yes" given below...!!  Askiitians_Expert Yagyadutt
9 years ago

Dear Christopher !

for k n=1 and for L n=2

Energy at every level is equal to  -13.6/n^2

So at n=1 it is -13.6 eV

and at n=2 it is -3.4 ev

differnece in the energy at two levels is  = 10.2 ev which is definately a negative...means if an electron jumps from L to K it will release energy ..

So wavelength corresponding to 10.2 ev will be emitted ..

hc/lamda = 10.2 ev

12400/lamda = 10.2

or lamda = 12400/10.2   A   (calculate for the answer)

regards

Yagya