Saved Bookmarks
| 1. |
The radius of hydrogen atom in its ground state is 5.3xx10^(-11)m. After collision with an electron it is found to have a radius 21.2xx10^(-11) m, then principle quantum number n of the final state of the atom ...... |
|
Answer» Solution :`rpropn^(2)` `(r_(2))/(r_(1))=((n_(2))/(n_(1)))^(2)` `r_(2)=` radius in final state `r_(1)=` radius in ground state `:.(21.2xx10^(-11))/(5.3xx10^(-11))=((n_(2))/(1))^(2)` `:.4=n_(2)^(2)` `:.n_(2)=2` |
|