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Question Number 31193 by Tinkutara last updated on 03/Mar/18

Answered by ajfour last updated on 03/Mar/18

E_n =−(e^2 /(4πε_0 (2r)))+m_e v^2 ((m_e v^2 )/r)=(e^2 /(4πε_0 (2r)^2 )) ⇒ m_e v^2 =(1/2)((e^2 /(4πε_0 (2r)))) ...(i) ⇒ E_n =−(1/2)×(e^2 /(4πε_0 (2r))) m_e vr=((nh)/(2π)) ⇒ m_e ^2 v^2 r^2 =((n^2 h^2 )/(4π^2 )) ..(ii) (ii)÷(i) gives m_e r^2 = ((n^2 h^2 )/(4π^2 ))(((16πε_0 r)/e^2 )) or r=((4n^2 h^2 ε_0 )/(πm_e e^2 )) ⇒ E_n =−(e^2 /(16πε_0 ))(((πm_e e^2 )/(4n^2 h^2 ε_0 ))) △E = ((hc)/λ) = ((m_e e^4 )/(64ε_0 ^2 h^2 ))((1/n^2 )−(1/m^2 )) ⇒ (1/λ)= ((m_e e^4 )/(64ε_0 ^2 ch^3 ))((1/n^2 )−(1/m^2 )) Hence R_p =((me^4 )/(64𝛆_0 ^2 ch^3 )) For hydrogen R_H =((me^4 )/(8ε_0 ^2 ch^3 )) ⇒ R_p = (R_H /8) .

$${E}_{{n}} =−\frac{{e}^{\mathrm{2}} }{\mathrm{4}\pi\epsilon_{\mathrm{0}} \left(\mathrm{2}{r}\right)}+{m}_{{e}} {v}^{\mathrm{2}} \\ $$$$\frac{{m}_{{e}} {v}^{\mathrm{2}} }{{r}}=\frac{{e}^{\mathrm{2}} }{\mathrm{4}\pi\epsilon_{\mathrm{0}} \left(\mathrm{2}{r}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:{m}_{{e}} {v}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{e}^{\mathrm{2}} }{\mathrm{4}\pi\epsilon_{\mathrm{0}} \left(\mathrm{2}{r}\right)}\right)\:\:\:\:...\left({i}\right) \\ $$$$\Rightarrow\:\:{E}_{{n}} =−\frac{\mathrm{1}}{\mathrm{2}}×\frac{{e}^{\mathrm{2}} }{\mathrm{4}\pi\epsilon_{\mathrm{0}} \left(\mathrm{2}{r}\right)} \\ $$$${m}_{{e}} {vr}=\frac{{nh}}{\mathrm{2}\pi}\:\:\Rightarrow\:\:{m}_{{e}} ^{\mathrm{2}} {v}^{\mathrm{2}} {r}^{\mathrm{2}} =\frac{{n}^{\mathrm{2}} {h}^{\mathrm{2}} }{\mathrm{4}\pi^{\mathrm{2}} }\:\:\:..\left({ii}\right) \\ $$$$\left({ii}\right)\boldsymbol{\div}\left({i}\right)\:\:{gives} \\ $$$$\:\:{m}_{{e}} {r}^{\mathrm{2}} =\:\frac{{n}^{\mathrm{2}} {h}^{\mathrm{2}} }{\mathrm{4}\pi^{\mathrm{2}} }\left(\frac{\mathrm{16}\pi\epsilon_{\mathrm{0}} {r}}{{e}^{\mathrm{2}} }\right) \\ $$$${or}\:\:\:\:\:{r}=\frac{\mathrm{4}{n}^{\mathrm{2}} {h}^{\mathrm{2}} \epsilon_{\mathrm{0}} }{\pi{m}_{{e}} {e}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\:{E}_{{n}} =−\frac{{e}^{\mathrm{2}} }{\mathrm{16}\pi\epsilon_{\mathrm{0}} }\left(\frac{\pi{m}_{{e}} {e}^{\mathrm{2}} }{\mathrm{4}{n}^{\mathrm{2}} {h}^{\mathrm{2}} \epsilon_{\mathrm{0}} }\right) \\ $$$$\bigtriangleup{E}\:=\:\frac{{hc}}{\lambda}\:=\:\frac{{m}_{{e}} {e}^{\mathrm{4}} }{\mathrm{64}\epsilon_{\mathrm{0}} ^{\mathrm{2}} {h}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{{m}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\:\:\frac{\mathrm{1}}{\lambda}=\:\frac{{m}_{{e}} {e}^{\mathrm{4}} }{\mathrm{64}\epsilon_{\mathrm{0}} ^{\mathrm{2}} {ch}^{\mathrm{3}} }\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{{m}^{\mathrm{2}} }\right) \\ $$$${Hence}\:{R}_{{p}} =\frac{\boldsymbol{{me}}^{\mathrm{4}} }{\mathrm{64}\boldsymbol{\epsilon}_{\mathrm{0}} ^{\mathrm{2}} \boldsymbol{{ch}}^{\mathrm{3}} } \\ $$$${For}\:{hydrogen}\:\:{R}_{{H}} =\frac{{me}^{\mathrm{4}} }{\mathrm{8}\epsilon_{\mathrm{0}} ^{\mathrm{2}} {ch}^{\mathrm{3}} } \\ $$$$\Rightarrow\:\:\:{R}_{{p}} =\:\frac{{R}_{{H}} }{\mathrm{8}}\:\:. \\ $$$$ \\ $$

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