Show that Equation 4-19 for the radius of the first Bohr orbit and Equation 4-20 for the magnitude of the lowest energy for the hydrogen atom can be written as
a_{0}=\frac{h c}{\alpha m c^{2}}=\frac{\lambda_{c}}{2 \pi \alpha} \quad E_{1}=\frac{1}{2} \alpha^{2} m c^{2}
where \lambda_{c}=h / m c is the Compton wavelength of the electron and \alpha=k e^{2} / h c is the fine-structure constant. Use these expressions to check the numerical values of the constants a_{n} and E_{1}.

Passo 1

O enunciado pede para mostrar que as fórmulas clássicas da órbita de Bohr,
a_{0}=\frac{4\pi\epsilon_{0}\hbar^{2}}{m e^{2}},
e da energia do estado fundamental,
E_{1}=-\frac{m}{2\hbar^{2}}\Bigl(\frac{e^{2}}{4\pi\epsilon_{0}}\Bigr)^{2},
podem ser reescritas em termos do comprimento de Compton \lambda_{c}=h/(m c) e da constante de estrutura fina \alpha=k\,e^{2}/(\hbar c). Em seguida, usar essas formas para calcular numericamente a_{0}, a_{n}=n^{2}a_{0} e |E_{1}|.

Passo 2

Partimos de
a_{0}=\frac{4\pi\epsilon_{0}\hbar^{2}}{m e^{2}}.
Como k=1/(4\pi\epsilon_{0}), temos
\frac{e^{2}}{4\pi\epsilon_{0}}=k\,e^{2}.
E pela definição de \alpha,
k\,e^{2}=\alpha\,\hbar\,c.
Logo
a_{0} =\frac{\hbar^{2}}{m\,(e^{2}/4\pi\epsilon_{0})} =\frac{\hbar^{2}}{m\,\alpha\,\hbar\,c} =\frac{\hbar}{m\,c\,\alpha}.

Passo 3

Usando \hbar=h/(2\pi) e \lambda_{c}=h/(m c), obtemos
a_{0} =\frac{h}{2\pi\,m\,c\,\alpha} =\frac{\lambda_{c}}{2\pi\,\alpha}.
Multiplicando numerador e denominador por c dá também
a_{0}=\frac{h\,c}{\alpha\,m\,c^{2}}.

Passo 4

O valor do estado fundamental é
E_{1} =-\frac{m}{2\hbar^{2}}\Bigl(\frac{e^{2}}{4\pi\epsilon_{0}}\Bigr)^{2} =-\frac{m}{2\hbar^{2}}(\alpha\,\hbar\,c)^{2} =-\frac12\,\alpha^{2}\,m\,c^{2}.
A magnitude fica
|E_{1}|=\frac12\,\alpha^{2}\,m\,c^{2}.

Passo 5

Vamos aos valores numéricos:
1. \alpha\approx7,297\times10^{-3}.
2. \lambda_{c}=\frac{h}{m c}\approx2,426\times10^{-12}\,\text{m}.
3. Então
a_{0}=\frac{\lambda_{c}}{2\pi\,\alpha} \approx\frac{2,426\times10^{-12}}{2\pi\,(7,297\times10^{-3})} \approx5,29\times10^{-11}\,\text{m}.
Logo
a_{n}=n^{2}a_{0}\approx n^{2}\times5,29\times10^{-11}\,\text{m}.
4. E
|E_{1}|=\frac12\,\alpha^{2}\,m\,c^{2} \approx\frac12\,(7,297\times10^{-3})^{2}\times511\,\text{keV} \approx13,6\,\text{eV}.

Resposta

a_{n}=n^{2}\,(5,29\times10^{-11}\,\text{m})
|E_{1}|=13,6\,\text{eV}

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