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 | 28th Feb, 2012, 01:57: PM

Expert Answer:

lamda = wavelength
1/(lamda) = R(1/nf2 -1/ni2)
nf  is 2 in He +  for shortest wavelength
ni  is infinity
nf  is 3 in He +  for longest wavelength

Answered by  | 2nd Mar, 2012, 10:28: AM

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