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

Short Answer

Expert verified

Wavelength of the photons corresponding to transitions 43,84,94in the given range are , and respectively.

Step by step solution

01

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 know that, energy of a photon
E=hcλ

Where, his Planck's constant, cis the speed of light, and localid="1659382795937" λwavelength of photon emitted.

Here, the numerical value of is given by,

hc=1240eV.nm

Define the energy atlocalid="1659618186089" 450nm as below.
E=hcλ=1240eV.nm450nm=2.76eV

Define the energy at450 as below.

E=hcλ=1240eV.nm500nm=2.49eV

02

Energies of electron in different orbits:

As you know that Energy of an electron in ‘n’ th orbit is given by,
E=-z213.6eVn2(n=1,2,3............)


Where, zis the atomic number of hydrogen-like atom andn is the principal quantum number.

En=-2213.6eVn2
En=-54.4eVn2 .......(1)
Now, using equation (1), you get.

E1=-54.4eV,E2=-13.6eV,E3=6.04eV,E4=3.4eV,E5=2.18eV,E6=1.51eV.E7=1.11eV,E8=0.85eV,E9=0.67eV,E10=0.54eV,...


03

Energies transmitted in the range 2.76 eV and 2.49 eV:

From step 2, you get,

The transitions43.84.94will correspond to energieslocalid="1659618569980" 2.64eV, 2.55eVand 2.73eVwhich are in the given range.

04

Wavelength of the transmissions:

ifE=hcλ

For432.64eV=1240eV.nmλλ=470nm


localid="1659381996658" For842.55eV=1240eV.nmλ
λ=487nm




For942.73eV=1240eV.nmλλ=455nm

05

Conclusion:

Wavelengths of the photons corresponding to transitions 43.84.94in the given range are470nm, 487nm and 455 nm respectively.

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