Question

Exercise 81 obtained formulas for hydrogen like atoms in which the nucleus is not assumed infinite, as in the chapter, but is of mass, ${\mathit{m}}_{\mathbf{1}}$while ${\mathit{m}}_2$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?

Short answer

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

(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.

If$\mathit{\mu }$is the effective mass and m is the mass, the actual energy will be $\mathit{\mu }\mathbf{/}\mathit{m}$times the hydrogen energy.

Now, if $m_e$and $m_p$are the mass of electron and proton,

$\begin{array}{rcl}\frac{\mu }{m}& =& \frac{m_e{m}_p/(m_e+m_p)}{m_e}\\ & =& \frac{1}{1+m_e/m_p}\\ & =& \frac{1}{1+9.11\times {10}^{31}/1.673\times {10}^{-27}}\\ & =& 1-0.00054\end{array}$

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

(b) Ground state energy of Deuterium and Hydrogen:

Also, the ratio of energies $\left(E_n\right)$ of Deuterium and Hydrogen will be equal to the ratio of their reduced masses $\left(\mu \right)$.

$$\begin{gathered} \begin{array}{rcl}\frac{E_{Deuterinum}}{E_{Hydrogen}}& =& \frac{{\mu }_{Deuterinum}}{{\mu }_{Hydrogen}}\\ & =& \frac{m_e{m}_{Deut}/\left(m_e{m}_{Deut}\right)}{m_e{m}_p/(m_e/m_p)}\\ & =& \frac{1+(m_e/m_p)}{1+(m_e/m_{Deut})}\\ & & \\ \frac{E_{Deuterium}}{E_{Hydrogen}}& =& \frac{1+(9.11\times {10}^{-31}/1.673\times {10}^{-27})}{1+(9.11\times {10}^{-31}/2\times 1.673\times {10}^{-27}) \\ =1.00027}\end{array} \end{gathered}$$

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