Question

Self-test 9A.3 The emission spectrum of atomic deuterium shows lines at 15 238, 20 571, 23 039, and 24 380 cm−1, which correspond to transitions to the same lower state. Determine (a) the ionization energy of the lower state, (b) the ionization energy of the ground state, (c) the mass of the deuteron (by expressing the Rydberg constant in terms of the reduced mass of the electron and the deuteron, and solving for the mass of the deuteron).

Answer: (a) 328.1 kJ mol−1, (b) 1312.4 kJ mol−1, (c) 2.8 × 10−27 kg, a result very sensitive to

Answer 1

a)The spectroscopic determination of ionization energy depends on the series limit, that is ,at the wavenumber at which the series terminates and becomes a continuum.

Say,if the upper state lies at an energy -hcRH/n2,then for a transition to lower state Elower, the photon of wavenumber v, is emitted .

V=-RH/n2-Elower/hc

-Elower=I=ionization energy

So, V=-RH/n2+I/hc

A plot of V vs 1/n2 gives a slope=-RH and intercept=I/hc

V= 15 238, 20 571, 23 039, and 24 380 cm−1

As n is not specified so ,it is obtained by making several choices ,and selecting one that can give a straight line.Slope=RH=109679 cm-1

For deuterium (2e ) n could be 2,3,4,5,6 …as lowest is n=1 ground state

1/n2=0.25, 0.11,0.0625,0.04,0.03,0.02

n=3 gives straight line

plot

intercept=I/hc=26000 cm-1

I=26000 cm-1 *hc=26000 cm-1*10^2 m-1/cm-1* 6.63*10^-34 m2kg/s *3.00*10^8 m/s=517.14*10^-21 J

I/mole=517.14*10^-21 J *6.022*10^23 mol-1=311.4 KJ/mol

b) for ground state of hydrogen, transition is from n=infinity to n=1

I=hcRH    [from equation, E=hcv=RH(1/n1 – 1/n2) n1=infinity ,n2=1]

Slope=15.238-20.571/0.0625-0.04=5.333/1.56=3.41*103=3418.58 cm-1

I=6.63*10^-34 m2kg/s *3.00*10^8 m/s *3418.58 cm-1 *10^2 m-1/cm-1=67995.552*10^-24 J/molecule

I/mole=67995.552*10^-24 J/molecule *6.022*10^23 mol-1=409469.23*10-1=40946.9J /mol

c)RD=MD/me *R

RD=rydberg constant for deuterium=1.0967*10^7 m-1

Me=mass of electron=9.10938*10^-31 kg

R=rydberg constant =109679*10^2 m-1

MD=RD/R*me= reduced mass of deuterium=9.1069*10^-31 kg

So MD=mD*me/mD+me=9.1069*10^-31 kg=mD*9.10938*10^-31 kg /mD+9.10938*10^-31 kg

mD=9.1069*10^-31 kg