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?
The radii = 5.29x10^-11*n^2
So i did (5.29e-11)(81)= 4.2849e-9. But it still says that is incorrect...
are you certain n is supposed to be 9? That is pretty far out, I would expect in Be+3 an n=3 radii.
That is what the question says n is
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
To find the radius of the triply ionized beryllium atom (Be3+) when it is in the n = 9 excited state, we can use the Bohr model of the atom.
In the Bohr model, the radius of an electron's orbit in the nth energy level is given by the formula:
r = (n^2 * h^2) / (4π^2 * m * Z * e^2)
where:
- r is the radius of the orbit
- n is the principal quantum number of the energy level
- h is the Planck constant (6.62607015 × 10^-34 m^2 kg / s)
- m is the reduced mass of the electron-nucleus system (in this case, it is essentially the electron mass)
- Z is the atomic number of the nucleus (in this case, Z = 4 for beryllium)
- e is the elementary charge (1.602176634 × 10^-19 C)
Using this formula, we can calculate the radius of the Be3+ ion in the n = 9 excited state.
Plugging in the values:
n = 9
h = 6.62607015 × 10^-34 m^2 kg / s
m = mass of an electron (9.10938356 × 10^-31 kg)
Z = 4
e = 1.602176634 × 10^-19 C
Calculating:
r = (9^2 * (6.62607015 × 10^-34)^2) / (4π^2 * (9.10938356 × 10^-31) * 4 * (1.602176634 × 10^-19)^2)
Upon evaluating the above expression, the radius of the triply ionized beryllium ion (Be3+) in the n = 9 excited state can be obtained.