Tue February 28, 2012 By: Sudhi 15477

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+.

Expert Reply
Fri March 02, 2012
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
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