Asked by DCM
A triply ionized beryllium atom (Be3+) has only one electron in orbit about the nucleus. What is the radius of the ion when it is in the n = 9 excited state?
Answers
Answered by
bobpursley
The radii = 5.29x10^-11*n^2
Answered by
DCM
So i did (5.29e-11)(81)= 4.2849e-9. But it still says that is incorrect...
Answered by
bobpursley
are you certain n is supposed to be 9? That is pretty far out, I would expect in Be+3 an n=3 radii.
Answered by
DCM
That is what the question says n is
Answered by
Elena
For hydrogen-like ion
r(n) = 0.529•10^-10•n^2/Z ,
where n is the NUMBER of exited state,
and Z is the number of the element in the Periodic Table
(for Be Z =4)
=0.529•10^-10•81/4 = 1.071•10^-9 m
r(n) = 0.529•10^-10•n^2/Z ,
where n is the NUMBER of exited state,
and Z is the number of the element in the Periodic Table
(for Be Z =4)
=0.529•10^-10•81/4 = 1.071•10^-9 m
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