What is the wavelength of light that causes an electronic transition for ${Li}^{2 +}$ $"Li"^(2+)$ if $n_{f} - n_{i} = 2$ $n_f - n_i = 2$ and $n_{i} + n_{f} = 4$ $n_i + n_f = 4$ ?

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What is the wavelength of light that causes an electronic transition for ${Li}^{2 +}$ $"Li"^(2+)$ if $n_{f} - n_{i} = 2$ $n_f - n_i = 2$ and $n_{i} + n_{f} = 4$ $n_i + n_f = 4$ ?

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Answer 1 · 2017-08-08T01:34:29

I got $11.39 nm$ $"11.39 nm"$ , for a change in energy of $108.88 eV$ $"108.88 eV"$ .

Well, we first dissect the minor puzzle here... Your system of equations is:

$n_{f} - n_{i} = 2$ $n_f - n_i = 2$

$n_{i} + n_{f} = 4$ $n_i + n_f = 4$

where:

$n_{f}$ $n_f$ is your destination energy level in a hydrogen-like atom.

$n_{i}$ $n_i$ is the energy level from which the electron started transitioning in that hydrogen-like atom.

By simple substitution,

$n_{f} = 2 + n_{i}$ $n_f = 2 + n_i$

$\Rightarrow n_{i} + 2 + n_{i} = 4$ $=> n_i + 2 + n_i = 4$

$=> ul(n_i = 1)$

$=> ul(n_f = 3)$

Indeed, $3 - 1 = 2$ and $1 + 3 = 4$ . Anyways...

Now that we know the electronic transition is $color(green)(n = 1 -> 3)$ , we invoke the Rydberg equation for hydrogen-like atoms, i.e. $"He"^(+)$ , $"Li"^(2+)$ , $"Be"^(3+)$ , etc.:

$bb(DeltaE = -Z^2 cdot "13.61 eV"(1/n_f^2 - 1/n_i^2))$

where:

$DeltaE$ is the change in energy due to the transition.

$-"13.61 eV"$ is the ground-state energy of hydrogen atom. The magnitude, $R_H ~~ "13.61 eV"$ , is also called the Rydberg constant.

$Z$ is the atomic number.

The wavelength $lambda$ is gotten by converting the resultant change in energy, which corresponds to the energy of the emitted photon during the electronic relaxation. In other words:

![https://upload.wikimedia.org/]( useruploads.socratic.org )

$bb(DeltaE = E_"photon" = hnu = (hc)/(lambda))$

where:

$h = 6.626 xx 10^(-34) "J"cdot"s"$ is Planck's constant.

$c = 2.998 xx 10^(8) "m/s"$ is the speed of light.

$nu$ is the frequency of the photon in $"s"^(-1)$ .

And so, the Rydberg equation for wavelength is:

$bb((DeltaE)/(hc) = 1/lambda = -(Z^2 cdot "13.61 eV")/(hc)(1/n_f^2 - 1/n_i^2))$

Knowing that, the wavelength will be:

$lambda = [-(3^2 cdot 13.61 cancel"eV" xx (1.602 xx 10^(-19) "J")/(cancel"1 eV"))/((6.626 xx 10^(-34) "J"cdot"s" xx 2.998 xx 10^(8) "m/s"))(1/3^2 - 1/1^2)]^(-1)$

$=$ $1.139 xx 10^(-8) "m"$

In nicer units, we then have:

$color(blue)(lambda) = 1.139 xx 10^(-8) cancel"m" xx (10^9 "nm")/(cancel"1 m")$

$=$ $color(blue)(ul"11.39 nm")$

Just to be sure you get the right value, the energy you get, which is easier to calculate, would have been:

$color(blue)(DeltaE) = -Z^2 cdot "13.61 eV"(1/3^2 - 1/1^2)$

$= -3^2 cdot "13.61 eV"(1/9 - 1/1)$

$=$ $color(blue)(ul"108.88 eV")$

And to check, we look on NIST. The energy levels for $n = 1->3$ in $"Li"^(2+)$ are listed as:

[

These are in $"cm"^(-1)$ . Take the first one for example (ignore the slight differences in the two choices).

$lambda_(ref) = 1/("877919.12077 "cancel("cm"^(-1))) xx (10^7 "nm")/(cancel"1 cm")$

$=$ $"11.3906 nm"$ $~~$ $"11.39 nm"$

So, the wavelength we got is correct.

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What is the wavelength of light that causes an electronic transition for ${Li}^{2 +}$ $"Li"^(2+)$ if $n_{f} - n_{i} = 2$ $n_f - n_i = 2$ and $n_{i} + n_{f} = 4$ $n_i + n_f = 4$ ?

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What is the wavelength of light that causes an electronic transition for "Li"^(2+)Li2+"Li"^(2+) if n_f - n_i = 2nf−ni=2n_f - n_i = 2 and n_i + n_f = 4ni+nf=4n_i + n_f = 4?

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What is the wavelength of light that causes an electronic transition for ${Li}^{2 +}$ $"Li"^(2+)$ if $n_{f} - n_{i} = 2$ $n_f - n_i = 2$ and $n_{i} + n_{f} = 4$ $n_i + n_f = 4$ ?