An electron in He + ion is in a state having directional dependence and has a total of one node. A photon of frequency 9 5 RC causes this election to jump into a higher state which can have total fi

Question

An electron in He + ion is in a state having directional dependence and has a total of one node. A photon of frequency 9 5 RC ​ causes this election to jump into a higher state which can have total five possible orientations in space. Find magnitude of the change in ( n + ℓ ) values corresponding to two states? (R = Rydbergs constant, C = speed of light)

TestHub · Accepted Answer

Initial state: directional dependence (not s-orbital) and one node ( $n - 1 = 1$ ). So, $n = 2$ . Since it's not s-orbital, $ℓ \neq = 0$ . For $n = 2$ , $ℓ$ can be 0 or 1. Thus, $ℓ = 1$ (p-orbital). Initial state is 2p ( $n = 2, ℓ = 1$ ).

Energy absorbed: $E = h ν = h (\frac{5 RC}{9})$ .

For He⁺ ( $Z = 2$ ), the energy difference is given by:

$E = Z^{2} R c h (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$

$h (\frac{5 RC}{9}) = (2)^{2} R c h (\frac{1}{2 ^{2}} - \frac{1}{n _{2}^{2}})$

$\frac{5}{9} = 4 (\frac{1}{4} - \frac{1}{n _{2}^{2}})$

$\frac{5}{9} = 1 - \frac{4}{n _{2}^{2}}$

$\frac{4}{n _{2}^{2}} = 1 - \frac{5}{9} = \frac{4}{9}$

$n_{2}^{2} = 9 ⟹ n_{2} = 3$

Final state: $n = 3$ . It has five possible orientations in space, which means $2 ℓ + 1 = 5$ , so $ℓ = 2$ . This corresponds to a 3d orbital.

Initial $(n + ℓ)_{1} = 2 + 1 = 3$ .

Final $(n + ℓ)_{2} = 3 + 2 = 5$ .

Change in $(n + ℓ) = (n + ℓ)_{2} - (n + ℓ)_{1} = 5 - 3 = 2$ .

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