Question

Which electron transitions in singly ionized helium yield photon in the 450 - 500 nm(blue) portion of the visible range, and what are their wavelengths?

Short answer

Wavelength of the photons corresponding to transitions $4\to 3,8\to 4,9\to 4$in the given range are 470 nm, 487 nm, and 455 nm respectively.

Step-by-step solution

Energies in the range 450- 500 nm :

When an electron transitions from a higher orbit to the lower orbit, it loses energy in the form of photons. And the energy of that photon is equal to the difference in energy of the transition states in which the transition is observed.

As you knoww that, energy of a photon

$E=\frac{hc}{\lambda }$

Where, his Planck's constant, cis the speed of light, and$\lambda$

wavelength of photon emitted.

Here, the numerical value of is given by,

$hc=1240eV.nm$

Define the energy at $450nm$as below.

$$\begin{gathered} \begin{array}{rcl}E& =& \frac{hc}{\lambda }\\ & =& \frac{1240eV.nm}{450nm \\ =2.76eV \\ }\end{array} \end{gathered}$$

Define the energy at as below.

$$\begin{gathered} \begin{array}{rcl}E& =& \frac{hc}{\lambda }\\ & =& \frac{1240eV.nm}{500nm \\ =2.49eV}\end{array} \end{gathered}$$

Energies of electron in different orbits:

As you know that Energy of an electron in ‘n’ th orbit is given by,

$E=-z^2\frac{13.6eV}{n^2}(n=1,2,3............)$

Where, $z$is the atomic number of hydrogen-like atom and $n$is the principal quantum number.

$$\begin{gathered} E_n=-2^2\frac{13.6eV}{n^2 \\ } \end{gathered}$$

$E_n=-\frac{54.4eV}{n^2}$ ...... (1)

Now, using equation (1), you get.

$$\begin{gathered} E_1=-54.4eV,E_2=-13.6eV,E_3=6.04eV,E_4=3.4eV,E_5=2.18eV,E_6=1.51eV. \\ {E}_7=1.11eV,E_8=0.85eV,E_9=0.67eV,E_{10}=0.54eV,... \end{gathered}$$

Energies transmitted in the range 2.76 eV and 2.49 eV:

From step 2, you get,

The transitions$4\to 3.8\to 4.9\to 4$will correspond to energies$2.64eV$,$2.55eV$, and$2.73eV$which are in the given range.

Wavelength of the transmissions:

$ifE=\frac{hc}{\lambda }$

$$\begin{gathered} For4\to 3 \\ 2.64eV=\frac{1240eV.nm}{\lambda } \\ \lambda =470nm \end{gathered}$$

$$\begin{gathered} For8\to 4 \\ 2.55eV=\frac{1240eV.nm}{\lambda \\ } \end{gathered}$$

$\lambda =487nm$

role="math" localid="1659619345623" $$\begin{gathered} For9\to 4 \\ 2.73eV=\frac{1240eV.nm}{\lambda } \\ \lambda =455nm \end{gathered}$$

Conclusion:

Wavelengths of the photons corresponding to transitions $4\to 3.8\to 4.9\to 4$ in the given range are $470nm$, 487nm and 455 nm respectively.