Hydrogen gas absorbs light of wavelength 103 nm. Afterward, what wavelengths are
ID: 1978674 • Letter: H
Question
Hydrogen gas absorbs light of wavelength 103 nm.
Afterward, what wavelengths are seen in the emission spectrum? (answer in nm)
Explanation / Answer
Energy of the absorbed photon, E = hc/? =(6.63x10-34)*(3x108)/(103x10-9) J= 1.93x10-18 J or(1.93x10-18)/(1.6x10-19) eV = 12.07 eV Hydrogen atom on absorbing this photon makes a transition to ahigher state 'n' with an energy (-13.6/n2) from theground state with an energy (-13.6/12). This is shown asfollows: [(-13.6/n2) - (-13.6/12)] = 12.07 => 13.6/n2 = 13.6 - 12.07 => n2 = 13.6/1.53 = 8.89 and we get, n = 2.98 ˜ 3 Now, this excited hydrogen atom can emit three wavelengthscorresponding to three transitions: (i) n=3 to n=2 hc/? = [(-13.6/32) - (-13.6/22)] eV => (6.63x10-34)*(3x108)/? =13.6*[1/22 - 1/32]*1.6x10-19 J which gives, ? =(6.63x10-34)*(3x108)/(1.89*1.6x10-19)˜ 658 nm (ii) n=2 to n=1 hc/? = [(-13.6/22) - (-13.6/12)] eV => (6.63x10-34)*(3x108)/? =13.6*[1/12 - 1/22]*1.6x10-19 J which gives, ? =(6.63x10-34)*(3x108)/(10.2*1.6x10-19)˜ 122 nm (iii) n=3 to n=1 hc/? = [(-13.6/32) - (-13.6/12)] eV => (6.63x10-34)*(3x108)/? =13.6*[1/12 - 1/32]*1.6x10-19 J which gives, ? =(6.63x10-34)*(3x108)/(12.09*1.6x10-19)˜ 103 nm
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