Question Number 50451 by JDlix last updated on 16/Dec/18
$${how}\:{can}\:{i}\:{derive}\:{a}\:{formula}\:{to}\:{calculate}\:{the}\:{speed}\:{of}\:{an}\:{electron} \\ $$$${in}\:{n}^{{th}} \:{orbit}\:{of}\:{a}\:{hydrogen}\:{atom}? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Dec/18
![(1/(4πε_0 ))((e×e)/r_n ^2 )=((mv^2 )/r_n ) angular momentum=mvr_n =((nh)/(2π)) mv^2 r_n =ke^2 eqn 1 [k=(1/(4πε_0 ))] mvr_n =((nh)/(2π)) eqn 2 deviding eqn 1 by eqn 2 v=((ke^2 )/(nh))×2π v=(1/(4πε_0 ))×(e^2 /(nh))×2π=(e^2 /(2ε_0 h))×(1/n) pls check....](Q50457.png)
$$\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\frac{{e}×{e}}{{r}_{{n}} ^{\mathrm{2}} }=\frac{{mv}^{\mathrm{2}} }{{r}_{{n}} } \\ $$$${angular}\:{momentum}={mvr}_{{n}} =\frac{{nh}}{\mathrm{2}\pi} \\ $$$${mv}^{\mathrm{2}} {r}_{{n}} ={ke}^{\mathrm{2}} \:\:\:{eqn}\:\:\mathrm{1}\:\:\:\:\:\:\left[{k}=\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\right] \\ $$$${mvr}_{{n}} =\frac{{nh}}{\mathrm{2}\pi}\:\:\:\:\:\:{eqn}\:\mathrm{2} \\ $$$${deviding}\:{eqn}\:\mathrm{1}\:{by}\:{eqn}\:\mathrm{2} \\ $$$${v}=\frac{{ke}^{\mathrm{2}} }{{nh}}×\mathrm{2}\pi \\ $$$${v}=\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }×\frac{{e}^{\mathrm{2}} }{{nh}}×\mathrm{2}\pi=\frac{{e}^{\mathrm{2}} }{\mathrm{2}\epsilon_{\mathrm{0}} {h}}×\frac{\mathrm{1}}{{n}} \\ $$$${pls}\:{check}.... \\ $$$$ \\ $$
Commented by peter frank last updated on 16/Dec/18
$${your}\:{right}\:\:{i}\:{get}\:{the}\:{same}\: \\ $$$${result} \\ $$
Answered by peter frank last updated on 16/Dec/18
Commented by JDlix last updated on 17/Dec/18
$${i}\:{want}\:{to}\:{thank}\:{both}\:{of}\:{u} \\ $$