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Grade: 12th pass
the hydrogen like species Li+2 is in a spherical symmetric state Swith one radial node, upon absorbing the light the ion underges transition to a state S2 with one radial node and its energy is equal to ground state energy of hydrogen atom then find
1.state S1 is
  • 1s
  • 2s
  • 2p
  • 3s of state S1 in units of the grund state energy of hydrogen
  • 0.75
  • 1.50
  • 2.25
  • 4.50
3.orbital angular moment of state S2
  • 0
  • 1
  • 2
  • 3       
can u plz ans with clear exp
5 years ago

Answers : (3)

Bhaveen Juthani
32 Points
1.The answer is 2s.Since S1 is spherically symmetrical,it must be an s orbital and since it has only one radial node,we can conclude that it a ‘2s’ orbital.(For an ‘ns’ orbital where n may be 1,2,3 etc. no. of radial nodes=n-1.)
2.The answer is 2.25.Please use E=(-Z2/n2)*13.6 where Z=Atomic no. and n=principal quantum no.-13.6 eV is the ground state energy of a H atom.
3.Total no. of nodes in an orbital with principal quantum no. n is (n-1).Total no. of angular nodes is the value of l.Now,according to the information given,using the above eqn. i could calculate that for state S2,n=3.Now since no. of radial nodes is given to be one the no. of angular nodes should be 2-1=1(Total no. of nodes=no. of angular nodes+no. of radial nodes).Since l=1 this is a 3p orbital.Could you please check if the options given are correct?The answer should be a multiple of h/2pi.
5 years ago
489 Points
Bhaveen  thanq
but plz check ur ans for 1
no. radial nodes=n-l-1
5 years ago
Bhaveen Juthani
32 Points
Hey Manyam...i know that the no. of radial nodes is general....but in my answer i had clearly written that considering the orital is an ‘s’ orbital(1s,2s,3s,etc.),then the no.of. radial nodes could be calculated using n-1.
I think the answer for no.1 should be 2s.Could you point out the mistake in my resaoning?
5 years ago
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