Saved Bookmarks
| 1. |
In the Bohr's model of hydrogen-like atom the force between the nucleus and the electron is modified as F=(e^(2))/(4piepsi_(0))((1)/(r^(2))+(beta)/(r^(3))), where beta is a constant. For this atom, the radius of the nth orbit in terms of the bohr radius (a_(0)=(epsi_(0)h^(2))/(mpie^(2))) is |
|
Answer» `r_(n)=a_(0)n-beta` From question, `F_(e)=(e^(2))/(4piepsi_(0))((1)/(r^(2))+(beta)/(r^(3)))`, where `beta`=constant  `F=(1)/(4piepsi_(0))(ZE^(2))/(r^(2))` We know that, `a_(n)=r_(n)=(epsi_(0)h^(2)n^(2))/(Zmpie^(2))` `a_(0)=r_(0)=(epsi_(0)h^(2))/(Zmpie^(2))` `therefore r_(n)=a_(n)n^(2)` As, we consider the nth electron which is attracted by nucleus while itself is repelled by the electron of inner shell to the consider shell. |
|