Question

In view ofProblem 5.1, we can correct for the motion of the nucleus in hydrogen by simply replacing the electron mass with the reduced mass.

(a) Find (to two significant digits) the percent error in the binding energy of hydrogen (Equation 4.77) introduced by our use of m instead of μ.

$E_1=-\left[\frac{m}{2\sout{h^2}}{\left(\frac{e^2}{4\mathrm{\pi }\infty \text{'}}\right)}^2\right]=-13.6eV$(4.77).

(b) Find the separation in wavelength between the red Balmer lines $\left(n=3\to n=2\right)$for hydrogen and deuterium (whose nucleus contains a neutron as well as the proton).

(c) Find the binding energy of positronium (in which the proton is replaced by a positron—positrons have the same mass as electrons, but opposite charge).

(d) Suppose you wanted to confirm the existence of muonic hydrogen, in which the electron is replaced by a muon (same charge, but 206.77 times heavier). Where (i.e. at what wavelength) would you look for the “Lyman-α” line $\left(n=2\to n=1\right)$?.

Short answer

(a) The percent error is 0.054%

(b)$\mathbf{∆}\mathit{\lambda }\mathbf{=}\frac{\mathbf{9}\mathbf{.}\mathbf{109}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{31}}}{\mathbf{2}\left(1.673\times {10}^{-27}\right)}\left(6.563\times {10}^{-7}\right)\mathit{m}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{79}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathit{m}\mathbf{.}$

(c)$\mathbf{\mu }\mathbf{=}\left(13.6/2\right)\mathbf{eV}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{8}\mathbf{eV}\mathbf{.}$

(d)$\mathit{\lambda }\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{54}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathit{m}\mathbf{.}$

Step-by-step solution

(a) Finding the percent error in the binding energy of hydrogen,

From Eq. 4.77,

$E_1=-\left[\frac{m_e}{2{\sout{h}}^2}{\left(\frac{e}{4\mathrm{\pi }\infty \text{'}}\right)}^2\right]=-13eV$ (4.77).

$E_1$is proportional to mass, so $\frac{△E_1}{E_1}=\frac{△m}{\mu }=\frac{m-\mu }{\mu }=\frac{m\left(m+M\right)}{mM}-\frac{M}{M}=\frac{m}{M}$

The fractional error is the ratio of the electron mass(m) to the proton mass(M):

$\frac{9.109\times {10}^{-31}kg}{1.673\times {10}^{-27}kg}=5.44\times {10}^{-4}$

. The percent error is 0.054% (pretty small).

(b) Finding the separation in wavelength between the red Balmer lines(n=3→n=2) 

From Eq. 4.94, R is proportional to m, so

$R=\frac{m}{4\mathrm{\pi c}{\sout{\mathrm{h}}}^3}{\left(\frac{e^2}{4\mathrm{\pi }\infty \text{'}}\right)}^2=1.097\times {10}^7{m}^{-1}$ (4.94).

$\frac{△\left(1+\lambda \right)}{\left(1+\lambda \right)}=\frac{△R}{R}=\frac{△\mu }{\mu }=-\frac{\left(1-{\lambda }^2\right)△\lambda }{\left(1-\lambda \right)}=-\frac{△\lambda }{\lambda }.$

So (in magnitude)role="math" localid="1656065390250" $△\lambda /\lambda =∆\mu /\mu .But\mu =mM/\left(m+M\right).$, where m = electron mass,

and M = nuclear mass.

$$\begin{gathered} △\mu =\frac{m\left(2m_p\right)}{m+2m_p}-\frac{m{m}_p}{m+m_p}=\frac{m{m}_p}{\left(m+m_p\right)\left(m+2m_p\right)}\left(2m+2m_p-m-2m_p\right). \\ =\frac{m^2{m}_p}{\left(m+m_p\right)\left(m+2m_p\right)}=\frac{m\mu }{m+2m_p} \\ \frac{△\lambda }{\lambda }=\frac{∆\mu }{\mu }=\frac{m}{m+2m_p}\approx \frac{m}{2m_p}so∆\lambda =\frac{m}{2m_p}{\lambda }_h,whereisthehydrogenwavelength. \\ \frac{1}{\lambda }=R\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36}R\Rightarrow \lambda =\frac{36}{5R}=\frac{36}{5\left(1.097\times {10}^7\right)}m=6.563\times {10}^{-7}m. \\ \therefore ∆\lambda =\frac{9.109\times {10}^{-31}}{2\left(1.673\times {10}^{-27}\right)}\left(6.563\times {10}^{-7}\right)m=1.79\times {10}^{-10}m. \end{gathered}$$

(c) Finding the binding energy of positronium,

$$\begin{gathered} \mu =\frac{mm}{m+m}=\frac{m}{2} \\ ,sotheenergyishalfwhatitwouldbeforhydrogen \\ :\left(13.6/2\right)eV=6.8eV. \end{gathered}$$

(d) The “Lyman-α” line (n=2→n=1)is at,

$$\begin{gathered} \mu =\frac{m_p{m}_{\mu }}{m_p{m}_{\mu }};R\infty \mu , \\ soRischangedbyafactor\frac{m_p{m}_{\mu }}{m_p+m_{\mu }}-\frac{m_p{m}_e}{m_p+m_e}=\frac{m_{\mu }\left(m_p{m}_e\right)}{m_e\left(m_p+m_{\mu }\right)},ascomparedwith \\ hydrogen. \\ Forhydrogen,1/\lambda =R\left(1-1/4\right)=\frac{3}{4}R\Rightarrow \lambda =4/\left[3\left(1.097\times {10}^7\right)\right]m=1.215\times {10}^{-7}m,and\lambda \infty 1/R \\ ,soformuonichydrogentheLyman-alphalineisat \\ \lambda =\frac{m_{\mu }\left(m_p{m}_e\right)}{m_e\left(m_p+m_{\mu }\right)}\left(1.215\times {10}^{-7}m\right)=\frac{1}{206.77}\frac{\left(1.673\times {10}^{-27}+206.77\times 9.109\times {10}^{-31}\right)}{\left(1.673\times {10}^{-27}+9.109\times {10}^{-31}\right)}\left(1.215\times {10}^{-7}m\right) \\ =6.54\times {10}^{-10}m. \end{gathered}$$