# Wavelength of the transition

A hydrogen-like ion has a nucleus of charge +Ze and a single electron outside this nucleus. The energy levels of these ions are -ZsquaredRH/n2 (where Z = atomic number). Calculate the wavelength of the transition from n=3 to n=2 for He+, a hydrogen-like ion. In what region of the spectrum does this emission occur?

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

E= h c R (Z ^2) / (n^2)

h= Planck's constant= 6.6261x 10 ^(- 34) J s

c=speed of light=3 x 10 ^ 8 m/s

R=Rydberg constant= 1.09678 x 10 ^ 7 (m^ -1)

hcR= 2.18021E-18 Joules

This can also be expressed in electron volts (eV)

1 e V=1.602 x 10 ^ -19 Joules

Thus we divide the energy in Joules by 1.602 x 10 ^ -19 to get the energy in eV

or

hcR= 13.61 eV

Or E=13.61 (Z^2)/(n^2)

For Helium Z= 2

Thus E=13.61*2^2 / n^2 = 54.44 / n^2 eV

delta E=54.44 *{ 1/(n1 ^2)- 1/(n2 ^2)} eV

For transition from n=3 to n=2

n1= 2

n2= 3

or delta E= 7.56 eV

delta E=h nu =h c / lambda = 7.56 eV= 1.2111E-18 Joules

where E-18 denotes 10^-18

or lambda =hc/delta E= 1.64134E-07 m

which is equal to 1641.34 Angstrom or 164.134 nm

This is in the ultraviolet region of the spectrum

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