91影视

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 $渭/m$times the hydrogen energy.

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

Hence, the energy predicted by ignoring the proton鈥檚 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 $(渭)$.
localid="1659466466571" $$\begin{gathered} \frac{E_{Deuterinum}}{E_{Hydrogen}}=\frac{{渭}_{Deuterinum}}{{渭}_{Hydrogen}} \\ =\frac{m_e{m}_{Deut}/\left(m_e{m}_{Deut}\right)}{m_e{m}_p/(m_e/m_p)} \\ =\frac{1+(m_e/m_p)}{1+(m_e/m_{Deut})} \\ \frac{E_{Deuterium}}{E_{Hydrogen}}=\frac{1+(9.11脳{10}^{-31}/1.673脳{10}^{-27})}{1+(9.11脳{10}^{-31}/2脳1.673脳{10}^{-27}) \\ =1.00027} \end{gathered}$$

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