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?????? ?????????? ? ???????? ?? ???? ???? r ?? ??????? ????? ?? -ke^2÷3r^3 ?????? ???? ???? ??? ?? ???????? ?????? ?? ????? solve ???? ?? ??? ???? principal ?? use ????
Asked by aftab01561 | 02 Jul, 2022, 23:35: PM
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Force of attraction between electron and proton is same as centripetal force in imaginary atom .
 
Hence we have ,  begin mathsize 14px style m v squared over r space equals space K space fraction numerator e squared over denominator 3 r cubed end fraction end style
where m is mass of electron , v is speed of electron, r is orbital radius , K = 1/(4πεo ) is Coulomb's constant ,
e is charge magnitude of electron or proton .
 
From above expression, we get , Kinetic energy KE is given as
 
begin mathsize 14px style K E space equals space 1 half m space v squared space equals space K space fraction numerator e squared over denominator 6 space r squared end fraction end style
 
Potential energy PE is given as
 
begin mathsize 14px style P E space equals space minus integral subscript infinity superscript r F times d r apostrophe space equals space K over 2 integral subscript infinity superscript r fraction numerator e squared over denominator r apostrophe cubed end fraction d r apostrophe space equals space minus fraction numerator K e squared over denominator 6 r squared end fraction end style
Hence total energy = KE + PE = 0
 
Answered by Thiyagarajan K | 09 Jul, 2022, 21:17: PM
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