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Question: The hydrogen-like species \(L{i^{ + 2}}\) is in a spherically symmetric state \({S_1}\) with one rad...

The hydrogen-like species Li+2L{i^{ + 2}} is in a spherically symmetric state S1{S_1} with one radial node and its energy is equal to the ground state of the hydrogen atom.
The energy of the state S2{S_2} (next state) in units of the hydrogen atom ground state energy is:
A) 7.5
B) 1.50
C) 2.25
D) 4.50

Explanation

Solution

The ground state of Hydrogen atom is considered as S0{S_0} and the energy of Hydrogen atom is depicted as EH=−13.6Z2n2=−RZ2n2{E_H} = - 13.6\dfrac{{{Z^2}}}{{{n^2}}} = - R\dfrac{{{Z^2}}}{{{n^2}}}
In here Z is the atomic number and n is the no. of nodes. For Hydrogen atoms both Z and n will be unity itself. For finding the energies of Higher states like S1{S_1} , S2{S_2} , it can be found out by considering the energy of that state to be equal to the energy of the ground state S0{S_0} .

Complete answer: We have been given the information that Li+2L{i^{ + 2}} is in a spherically symmetric state. The only spherically symmetrical orbital is the s-orbital. The azimuthal quantum number l for s orbital is Zero. It has also been given that there is only a radial node present. The formula for finding the no. of radial nodes can be given as: r=n−l−1r = n - l - 1 -- (1)
Where r is the no. of radial nodes, n is the principal quantum number and l is the azimuthal quantum number. Therefore, for ground state S1{S_1} , the principal quantum number ‘n’ can be given as:
1=n−0−1→n=21 = n - 0 - 1 \to n = 2
If n is 2 and l is 0, we can say that the orbital is 2s. Hence S1{S_1} is in 2s orbital.
Firstly, let us find the ground state energy of Hydrogen Atom: EH=−R1212=−R{E_H} = - R\dfrac{{{1^2}}}{{{1^2}}} = - R -- (2) ( since the value of n and Z for Hydrogen atom is 1 )
The energy of S1{S_1} orbital can be given as: −RZ2n2 - R\dfrac{{{Z^2}}}{{{n^2}}}
Considering Z=3 for Li+2L{i^{ + 2}} and also that energy of S1{S_1} will be equal to the ground state energy of Hydrogen atom: −R1212=−R32n2→n=3 - R\dfrac{{{1^2}}}{{{1^2}}} = - R\dfrac{{{3^2}}}{{{n^2}}} \to n = 3
Note that it has only one radial node, therefore finding the value of l from the formula (1) given above; 1=3−l−1→l=11 = 3 - l - 1 \to l = 1
L will be 1 for p orbital, hence the orbital S2{S_2} will be 3p orbital.
Finding the energy of S2{S_2} orbital =−RZ2n2=−R3222=−94R=−2.25R = - R\dfrac{{{Z^2}}}{{{n^2}}} = - R\dfrac{{{3^2}}}{{{2^2}}} = - \dfrac{9}{4}R = - 2.25R-- (3)
Comparing equation (2) with (3) we get the energy of S2{S_2} orbital in terms of energy of Hydrogen atom.
Energy of S2{S_2} orbital =2.25EH = 2.25{E_H}
Hence the correct answer is Option C.

Note:
Note that the values of azimuthal quantum numbers for different orbitals have to be remembered. If l is zero then it is surely s-orbital. If l is 1 then it is p-orbital, if l is 2 then it is d orbital. If we know the number of radial nodes, we can use the formula r=n−l−1r = n - l - 1 to find the azimuthal quantum number.