Saved Bookmarks
| 1. |
The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10^(40). An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting, |
|
Answer» Solution :By thelawof gravition ,`(m_ev^2 )/(r )= (Gm _e m_p ) /(r^2)` `i.E.,m_e v^2 r= gm_e m_p `………(1) Bybohr.scondition`m_e vr =(nh )/( 2 pi)` `m_(e)^(2) V^2 r^2=(n^2 h^2)/( 4 pi ^2) `………(2) for ` 1^(ST)` ORBIT ,`n=1 , h=6.626xx 10^(-34) Js` ` G= 6.67 xx 10^(-11) Nm ^2 // kg ^2` ` m_e= 9xx 10^(-11) Kg , m_p= 1.67 xx 10^(-27 ) kg` Eq (2)`div `(1) ` impliesm_e r = (n^2 h^2)/( 4 pi ^2 Gm_e m_p)` ` r=(n^2 h^2)/( 4 pi ^2Gm_e^(2) m_p)` Substitingthesevaluesweget`r= 1.21 xx 10^(29)` m this value of .r. is much greaterthanthe SIZEOF the universe . |
|