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

Godfrey Classic Prince
633 Points
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

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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

Mukkamala Viswateja
29 Points
9 years ago

i think the wavelength is present in between the uv region.