11?????no one??????
Q→
A sample of H-atoms containing all atoms in some excited state is irradiated with light of wavelenth ∂ and atoms get reexcited to furthur higher orbit. on removing the light source , atoms de-excite to ground state with emission of radiations.the emission spectrum thus obtained consists of 10 spectral lines ,out of which 7 lines have wavelenth smaller than ∂ and 2 have wavelength larger than it.then determine that in which orbit were electrons of hydrogen atoms present initially(ie before irradiation with light of wavelength ∂) .
also if ∂=100nm then find the ionisation energy of this H-atom.
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10 Answers
11another very good question:
Q→
-->for examining emission spectrum of a H-like element, a sample of the same was irradiated with light of appropriate wavelength .the no of spectral lines observed on the screen was found to be=x.
In an another experiment same amount of sample of the same element was:
(i) put in a region of magnetic field in direction vertically downward(ie B=-B°j(vectorial representation))and this system was irradiated with light of same wavelength as was in initial case.this tym no of lines were found to be=y.
(ii) Put in magnetic field directing vertically upwaard (the magnitute of magnetic field being same as in case(i)) and this system was irradaited with light of same wavelength as in first two cases.this tym no of spectral lines were observed to be=z.
then which of following is true:
1.) x=y=z
2.) x>y>z
3.) x<y<z
4.) none of these is right
49the electron was initially in the 3rd orbit...
it got excited to 5 th orbit
and 1δ = K.{132-152}
we get value of K from here and
IE=hcδie=hcK.(112-1∞2)=hcK
K is famously known as the Rydberg constant
49x, y and z are the number of spectral lines observsed and they cannot change as orientation in magnetic field only splits the FINEl lines of spectrum due to presence od DEGENERATE orbitals...
according to me answer is x=y=z
11first one is right
second one's answer is
4)none of these is right
11subhu bhayiya have u done the Q1 using hit and trail??is there any other method ?
btw second question's relation wud be:
y=z>x(zeeman effect)
49no not hit and trial...its simple logic...
Look when we say that the no of spectral lines from the new excited state has 3 lines with δ less than applied δ to excite the electrons from old excited state to new excited state,
it means 3 lines have energy content more than the energy gap between the 2 excited states...
so these 3 lines are the spectral lines corresponding to the electron rebounding back to an orbital with n less than that of the old excited state...
so this brings me to this conclusion that if the electron resided in old excited state, it would give 3 spectral lines..
so the nC2=3, giving n=3
also the total no. of field lines in new excited state is 10 nC2=10 gives n=5.
note that:: nC2 can be written as n(n-1)2 (it is for the reference of those who do not know perm comb!)
so now as we have got n values 3 and 5 for the 2 excited state, the rest of the question can be solved as i mentioned before already!!
49@vaibhab: i had considered Zeeman effect..
but look the FINE spectrum was effected by zeeman effect.
i.e. bohr saw just discontinuous lines of spectrum,
but later when these were resolved, n=1 remained as 1 spectral line, n=2 was resolved into 2 lines n=3 resolved into 3 lines and et cetera...
now on exposure to magnetic field, these FINE spectral lines got splitted not the original lines of bohr...
and the question seemed that we are dealing with the lines bohr had seen...
so might be i got x=y=z
if the question is talking abt fine spectra then surely answer is x<y=z!!
cheers!!
11[7]
there's something wrong
i don't know if im making a mistake
but bhayiya where did i give that there are 3 lines with wavelength less than applied ∂
i gave (i mean the question gave[4]) that 7 lines have wavelength less than applied ∂ and 2 have greater wavelength
