Exercise 81 obtained formulas for hydrogen like atoms in which the nucleus is not assumed infinite, as in the chapter, but is of mass, m1while m2is the mass of the orbiting negative charge. (a) What percentage error is introduced in the hydrogen ground-state energy by assuming that the proton is of infinite mass? (b) Deuterium is a form of hydrogen in which a neutron joins the proton in the nucleus, making the nucleus twice as massive. Taking nuclear mass into account, by what percent do the ground-state energies of hydrogen and deuterium differ?

Short Answer

Expert verified

Answer:
(a) The energy predicted by ignoring the proton’s finite mass is too high by

(b) The ground state energy of hydrogen is less than 0.02% of the ground state energy of Deuterium.

Step by step solution

01

(a) Percentage error is introduced in the hydrogen ground-state energy:

Effective mass simplifies band structures because modelling the behaviour of a free particle with that mass can be observed easily.

Ifμis the effective mass and m is the mass, the actual energy will be μ/mtimes the hydrogen energy.

Now, if meand mpare the mass of electron and proton,

μm=memp/(me+mp)me=11+me/mp=11+9.11×1031/1.673×10-27=1-0.00054

Hence, the energy predicted by ignoring the proton’s finite mass is too high by 0.054% .

02

(b) Ground state energy of Deuterium and Hydrogen:

Also, the ratio of energies (En) of Deuterium and Hydrogen will be equal to the ratio of their reduced masses (μ).

EDeuterinumEHydrogen=μDeuterinumμHydrogen=memDeut/(memDeut)memp/(me/mp)=1+(me/mp)1+(me/mDeut)EDeuteriumEHydrogen=1+(9.11×10-31/1.673×10-27)1+(9.11×10-31/2×1.673×10-27)=1.00027

Hence, the ground state energy of hydrogen is less than 0.027% of the ground state energy of Deuterium.

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