the hydrogen like species Li⁺² is in a spherical symmetric state S₁ with 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

2.energy 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

thanq

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.

Bhaveen  thanq

but plz check ur ans for 1

no. radial nodes=n-l-1

Hey Manyam...i know that the no. of radial nodes is n-l-1...in 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?