Question 39.62: (a) What is the wavelength of light for the least energetic ......
(a) What is the wavelength of light for the least energetic photon emitted in the Balmer series of the hydrogen atom spectrum lines? (b) What is the wavelength of the series limit?
(a) The “home-base” energy level for the Balmer series is n = 2. Thus the transition with the least energetic photon is the one from the n = 3 level to the n = 2 level. The energy difference for this transition is
\Delta E=E_3-E_2=- (13.6 \,eV )\left(\frac{1}{3^2}-\frac{1}{2^2}\right)=1.889 \,eV .
Using hc = 1240 eV · nm, the corresponding wavelength is
\lambda=\frac{h c}{\Delta E}=\frac{1240 \, eV \cdot nm }{1.889 \, eV }=658 \,nm .
(b) For the series limit, the energy difference is
\Delta E=E_{\infty}-E_2=- ( 13.6\, eV) \left(\frac{1}{\infty^2}-\frac{1}{2^2}\right)=3.40 \,eV .
The corresponding wavelength is then \lambda=\frac{h c}{\Delta E}=\frac{1240 \,eV \cdot nm }{3.40\, eV }=366 \,nm .