(a) [tex]4.8\cdot 10^{-19} C[/tex]
The radius of the trajectory of a charged particle moving perpendicular to a magnetic field is given by
[tex]r=\frac{mv}{qB}[/tex]
where
m is the mass of the particle
q is its charge
v is its velocity
B is the strength of the magnetic field
In this problem, we have:
[tex]m=2.66\cdot 10^{-26} kg[/tex]
[tex]v=5.00\cdot 10^6 m/s[/tex]
B = 1.20 T
r = 0.231 m
Solving for q, we find its charge:
[tex]q=\frac{mv}{rB}=\frac{(2.66\cdot 10^{-26})(5.00\cdot 10^6)}{(0.231)(1.20)}=4.8\cdot 10^{-19} C[/tex]
(b) 3
The charge of an electron is
[tex]e=1.6\cdot 10^{-19}C[/tex]
While the charge of this oxygen ion is
[tex]q=4.8\cdot 10^{-19}C[/tex]
So, the ratio between the two charges is
[tex]\frac{q}{e}=\frac{4.8\cdot 10^{-19}}{1.6\cdot 10^{-19}}=3[/tex]
(c) Because an ion is an atom that has gained/lost an integer number of electrons
An ion is an atom that has gained/lost an integer number of electrons. In this particular case, we see that the charge of the oxygen ion is 3 times that of the electron: this means that the ion has gained/lost exactly 3 electrons.
The ratio found in part (b) cannot be a fraction, because that would mean that the atom has gained/lost a fractional number of electrons: but this is impossible, since the electron is a fundamental particle so it cannot be "divided", therefore the ratio must be an integer.
