Question

15.

Answer 1

(a) $4.8\cdot 10^{-19} C$

The radius of the trajectory of a charged particle moving perpendicular to a magnetic field is given by

$r=\frac{mv}{qB}$

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:

$m=2.66\cdot 10^{-26} kg$

$v=5.00\cdot 10^6 m/s$

B = 1.20 T

r = 0.231 m

Solving for q, we find its charge:

$q=\frac{mv}{rB}=\frac{(2.66\cdot 10^{-26})(5.00\cdot 10^6)}{(0.231)(1.20)}=4.8\cdot 10^{-19} C$

(b) 3

The charge of an electron is

$e=1.6\cdot 10^{-19}C$

While the charge of this oxygen ion is

$q=4.8\cdot 10^{-19}C$

So, the ratio between the two charges is

$\frac{q}{e}=\frac{4.8\cdot 10^{-19}}{1.6\cdot 10^{-19}}=3$

(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.