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

##### 1 Answer

I got

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

#n_f - n_i = 2#

#n_i + n_f = 4# where:

#n_f# is yourdestination energy levelin a hydrogen-like atom.#n_i# is the energy levelfrom whichthe electron started transitioning in that hydrogen-like atom.

By simple substitution,

#n_f = 2 + n_i#

#=> n_i + 2 + n_i = 4#

#=> ul(n_i = 1)#

#=> ul(n_f = 3)#

Indeed,

Now that we know the **electronic transition** is **Rydberg equation** for hydrogen-like atoms, i.e.

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

#DeltaE# is thechange in energydue to the transition.#-"13.61 eV"# is theground-state energyof hydrogen atom. The magnitude,#R_H ~~ "13.61 eV"# , is also called theRydberg constant.#Z# is theatomic number.

The **wavelength** **energy of the emitted photon** during the electronic relaxation. In other words:

#bb(DeltaE = E_"photon" = hnu = (hc)/(lambda))# where:

#h = 6.626 xx 10^(-34) "J"cdot"s"# isPlanck's constant.#c = 2.998 xx 10^(8) "m/s"# is thespeed of light.#nu# is thefrequencyof 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

These are in

#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.