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Home / Doubts and Solutions/JEE/Chemistry

the radius of an  orbital hydrogen atom is 0.85nm calculate velocity of electron in this orbital 

Asked by gargpuneet989 30th June 2018, 9:10 PM
Answered by Expert
Answer:
Given:
 
The radius of the 4th orbit of hydrogen atom = 0.85 nm
 
We space know comma For space hydrogen space atom comma space straight Z space equals space 1 mvr space equals fraction numerator nh over denominator 2 straight pi end fraction straight V space equals fraction numerator nh over denominator 2 straight pi open parentheses mr close parentheses end fraction space space space space space......... open parentheses 1 close parentheses straight r subscript straight n equals fraction numerator straight n squared straight h squared over denominator 4 straight pi squared kme squared end fraction space space space.......... open parentheses 2 close parentheses Substituting space eq space open parentheses 2 close parentheses space in space eq space open parentheses 1 close parentheses straight V equals space fraction numerator up diagonal strike nh 4 up diagonal strike straight pi straight k up diagonal strike straight m straight e squared over denominator up diagonal strike 2 straight pi end strike open parentheses up diagonal strike straight m straight n squared straight h squared close parentheses end fraction straight V equals fraction numerator space 2 πke to the power of straight e over denominator nh end fraction Where comma straight h equals space Plank apostrophe straight s space constant space space space equals space 6.626 cross times 10 to the power of negative 34 end exponent space Js straight n equals space orbit space number space equals space 4 straight m equals space mass space of space an space electron straight m space equals space 9.1 cross times 10 to the power of negative 31 end exponent straight e space equals charge space of space an space electron space space space space equals 1.6 cross times 10 to the power of negative 19 end exponent space straight C straight k space equals Coulomb apostrophe straight s space law space constant space space space equals space 9 cross times 10 to the power of 9 space Jm divided by straight C squared straight V space equals fraction numerator 2 cross times 3.14 cross times 9 cross times 10 to the power of 9 cross times open parentheses 1.6 cross times 10 to the power of negative 19 end exponent close parentheses squared over denominator space 6.626 cross times 10 to the power of negative 34 end exponent cross times 4 end fraction straight V space equals fraction numerator 144.69 cross times 10 to the power of negative 29 end exponent over denominator 26.05 cross times 10 to the power of negative 34 end exponent end fraction straight V space space space equals 5.55 cross times 10 to the power of 5 space straight m divided by sec
The velocity of an electron is 5.55×105 m/sec
Answered by Expert 2nd July 2018, 11:21 AM
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