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CBSE Class 12-science Answered

In an atom energy of electron in nth orbit is En = -13.6z2/n2ev, where zis atomic number. What is the shortest & longest wave length of emitted radiation n in singly ionized He+.
Asked by Sudheer Nair | 28 Feb, 2012, 01:57: PM
answered-by-expert Expert Answer
lamda = wavelength
1/(lamda) = R(1/nf2 -1/ni2)
nf  is 2 in He +  for shortest wavelength
ni  is infinity
and
nf  is 3 in He +  for longest wavelength
Answered by | 02 Mar, 2012, 10:28: AM
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