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

Answered by ajfour last updated on 03/Mar/18

$$E_n=-\frac{e^2}{4\pi {ϵ}_0\left(2r\right)}+m_e{v}^2$$$$\frac{m_e{v}^2}{r}=\frac{e^2}{4\pi {ϵ}_0{\left(2r\right)}^2}$$$$\Rightarrow m_e{v}^2=\frac{1}{2}\left(\frac{e^2}{4\pi {ϵ}_0\left(2r\right)}\right)...\left(i\right)$$$$\Rightarrow E_n=-\frac{1}{2}\times \frac{e^2}{4\pi {ϵ}_0\left(2r\right)}$$$$m_{e}vr=\frac{nh}{2\pi }\Rightarrow m_e^2{v}^2{r}^2=\frac{n^2{h}^2}{4{\pi }^2}..\left(ii\right)$$$$\left(ii\right)\mathbf{÷}\left(i\right)gives$$$$m_e{r}^2=\frac{n^2{h}^2}{4{\pi }^2}\left(\frac{16\pi {ϵ}_{0}r}{e^2}\right)$$$$orr=\frac{4n^2{h}^2{ϵ}_0}{\pi m_e{e}^2}$$$$\Rightarrow E_n=-\frac{e^2}{16\pi {ϵ}_0}\left(\frac{\pi m_e{e}^2}{4n^2{h}^2{ϵ}_0}\right)$$$$△E=\frac{hc}{\lambda }=\frac{m_e{e}^4}{64{ϵ}_0^2{h}^2}(\frac{1}{n^2}-\frac{1}{m^2})$$$$\Rightarrow \frac{1}{\lambda }=\frac{m_e{e}^4}{64{ϵ}_0^2{ch}^3}(\frac{1}{n^2}-\frac{1}{m^2})$$$$Hence{R}_p=\frac{{\mathit{m}\mathit{e}}^4}{64{\mathit{ϵ}}_0^2{\mathit{c}\mathit{h}}^3}$$$$Forhydrogen{R}_H=\frac{{me}^4}{8{ϵ}_0^2{ch}^3}$$$$\Rightarrow R_p=\frac{R_H}{8}.$$$$$$

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