Exercise 81 obtained formulas for hydrogen like atoms in which the nucleus is not assumed infinite, as in the chapter, but is of mass,$m_1$ while$m_2$is 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?

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

Ifis the effective mass and m is the mass, the actual energy will be $\mu /m$times the hydrogen energy.

Now, if $m_e$and$m_p$ are the mass of electron and proton,
$$\begin{gathered} \frac{\mu }{m}=\frac{m_e{m}_p/(m_e+m_p)}{m_e} \\ =\frac{1}{1+m_e/m_p} \\ =\frac{1}{1+9.11\times {10}^{31}/1.673\times {10}^{-27}} \\ =1-0.00054 \end{gathered}$$

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

Also, the ratio of energies( $E_n$)of Deuterium and Hydrogen will be equal to the ratio of their reduced masses $(\mu )$.
localid="1659466466571" $$\begin{gathered} \frac{E_{Deuterinum}}{E_{Hydrogen}}=\frac{{\mu }_{Deuterinum}}{{\mu }_{Hydrogen}} \\ =\frac{m_e{m}_{Deut}/\left(m_e{m}_{Deut}\right)}{m_e{m}_p/(m_e/m_p)} \\ =\frac{1+(m_e/m_p)}{1+(m_e/m_{Deut})} \\ \frac{E_{Deuterium}}{E_{Hydrogen}}=\frac{1+(9.11\times {10}^{-31}/1.673\times {10}^{-27})}{1+(9.11\times {10}^{-31}/2\times 1.673\times {10}^{-27}) \\ =1.00027} \end{gathered}$$

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