Question

(a) An oxygen\({\rm{ - 16}}\) ion with a mass of \({\rm{2}}{\rm{.66 \times 1}}{{\rm{0}}^{{\rm{ - 26}}}}{\rm{ kg}}\) travels at \({\rm{5}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{6}}}{\rm{ m/s}}\) perpendicular to a \({\rm{1}}{\rm{.20 - T}}\) magnetic field, which makes it move in a circular arc with a 0.231-m radius. What positive charge is on the ion? (b) What is the ratio of this charge to the charge of an electron? (c) Discuss why the ratio found in (b) should be an integer.

Short answer

1. The charge of the oxygen ion is obtained as: \({\rm{4}}{\rm{.8}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 19}}}}{\rm{ C}}\).
2. The ratio between the charge of the oxygen ion and the electron charge is obtained as:\({\rm{3}}\).
3. The ratio between the charge of the ion and the electron charge must be an integer just because the elements turn into ions by losing or gaining an electron or more. So, in this case, oxygen loses three electrons.

Step-by-step solution

Effect of magnetic field on the charged particle

When a charged particle moves in the presence of a magnetic field, a force acts on the mong charge particle given by Fleming’s right-hand rule and which deviates the path of the moving charged particle.

Charge on any ion

The electric charge on any ion is achieved after losing and/or gaining electrons, and this charge is always an integral multiple of the charge on one electron and known as quantization of the electric charge.

Evaluating the charge of an oxygen ion

(a)

To obtain the charge of an oxygen-16 ion.

The equation used is:\({\rm{r = }}\frac{{{\rm{mv}}}}{{{\rm{qB}}}}\).

Solving it for the value of\({\rm{q}}\)where the value of\({\rm{m}}\)is the ion’s mass.

\(\begin{aligned}{\rm{q}} &= \frac{{{\rm{mv}}}}{{{\rm{rB}}}}\\ &= \frac{{\left( {{\rm{2}}{\rm{.66 \times 1}}{{\rm{0}}^{{\rm{ - 26}}}}\;{\rm{kg}}} \right){\rm{ \times }}\left( {{\rm{5 \times 1}}{{\rm{0}}^{\rm{6}}}{\rm{ m/s}}} \right)}}{{\left( {{\rm{0}}{\rm{.231}}\;{\rm{m}}} \right){\rm{ \times }}\left( {{\rm{1}}{\rm{.2}}\;{\rm{T}}} \right)}}\\ &= {\rm{4}}{\rm{.8}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 19}}}}{\rm{ C}}\end{aligned}\)

Therefore, the charge of the oxygen ion is: \({\rm{4}}{\rm{.8}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 19}}}}{\rm{ C}}\).

Evaluating the ratio

(b)

To obtain the ratio of the evaluated charge to an electron. Dividing the charge we got by the electron's charge \({\rm{e}}\).

\(\begin{aligned}\frac{{\rm{q}}}{{\rm{e}}} &= \frac{{{\rm{4}}{\rm{.8 \times 1}}{{\rm{0}}^{{\rm{ - 19}}}}{\rm{ C}}}}{{{\rm{1}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 19}}}}{\rm{ C}}}}\\ &= {\rm{3}}\end{aligned}\)

Hence, the ratio between the charge of the oxygen ion and the electron charge is 3

Discuss why the ratio found in part b should be an integer?

(c)

The ratio found must surely be an integer as the elements turn into ions by losing or gaining an electron or more. Then, in this case, the oxygen has lost three electrons.