Question

A certain five cent coin contains\[{\rm{5}}{\rm{.00 g}}\]of nickel. What fraction of the nickel atoms’ electrons, removed and placed\[{\rm{1}}{\rm{.00 m}}\]above it, would support the weight of this coin? The atomic mass of nickel is\[{\rm{58}}{\rm{.7}}\], and each nickel atom contains\[{\rm{28}}\]electrons and\[{\rm{28}}\]protons.

Short answer

The fraction of electrons removed from the nickel atom and placed \(1.00{\rm{ m}}\) above it to support the weight the of five cent coin is \({\rm{1}}{\rm{.02}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 11}}}}\).

Step-by-step solution

Mole

The amount of substance that contains\({\rm{6}}{\rm{.022}} \times {\rm{1}}{{\rm{0}}^{{\rm{23}}}}\)elementary entities of the given substance is known as one mole. The number of moles of any substance can be calculated by taking the ratio of given mass of the substance to its molar mass.

\(N = \frac{m}{M}\)

Here, N is the number of moles, m is the mass, and M is the molar mass of the substance.

Total number of electrons

Number of moles of nickel is,

\(N = \frac{m}{M}\)

Here, m is the mass of five cent nickel coin \(\left( {m = 5.00{\rm{ }}g} \right)\), and M is the molar mass of the nickel \(\left( {M = 58.7{\rm{ }}g} \right)\)

Substituting all known values,

\(\begin{aligned} N &= \frac{{5.00{\rm{ }}g}}{{58.7{\rm{ }}g}}\\ &= 0.085\end{aligned}\)

Since, one mole of nickel contains \(6.022 \times {10^{23}}\)nickel atom. Therefore, total number of nickel atoms is,

\(\begin{aligned} {N_a} &= \left( {0.085} \right) \times \left( {6.022 \times {{10}^{23}}} \right)\\ &= 5.1187 \times {10^{22}}\end{aligned}\)

We know that one nickel atom contains 28 electrons. Therefore, total of electrons in 5 g of nickel five cent coin is,

\(\begin{aligned} {N_e} &= \left( {28} \right) \times \left( {5.1187 \times {{10}^{22}}} \right)\\ &= 1.433 \times {10^{24}}\end{aligned}\)

Number of  electrons removed

The weight of five cents of nickel coin is,

\({F_g} = mg\)

Here, m is the mass of five cents nickel coin \(\left( {m = 5{\rm{ }}g} \right)\), and g is the acceleration due to gravity \(\left( {g = 9.8{\rm{ m/}}{{\rm{s}}^{\rm{2}}}} \right)\).

Let n be the number of electrons removed from the coin. Therefore, the magnitude of net charge removed is,

\({q_1} = ne\)

Here, n is the number of electrons removed, and e is the fundamental unit of charge \(\left( {e = 1.6 \times {{10}^{ - 19}}{\rm{ }}C} \right)\).

The net charge gained by the five cents nickel coin is,

\({q_2} = ne\)

The electrostatic force between electrons removed and five cents nickel coin is,

\[{F_e} = \frac{{K{q_1}{q_2}}}{{{r^2}}}\]

Here, K is the electrostatic force constant \(\left( {K = 9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\), \({q_1}\) is the net charge removed from the nickel coin, \({q_2}\) is the net charge on the nickel coin, and r is the distance of separation \(\left( {r = 1.00{\rm{ }}m} \right)\).

From equation (1.2), (1.3), and (1.4),

\[{F_e} = \frac{{K{{\left( {ne} \right)}^2}}}{{{r^2}}}\]

Since, the weight of the five cents nickel coin is balanced by the electrostatic force.

Equating equation (1.1) and (1.5),

\(mg = \frac{{K{{\left( {ne} \right)}^2}}}{{{r^2}}}\)

The expression for the number of electrons is,

\(n = \sqrt {\frac{{mg{r^2}}}{{K{e^2}}}} \)

Substituting all known values,

\(\begin{aligned} n &= \sqrt {\frac{{\left( {5.00{\rm{ }}g} \right) \times \left( {9.8{\rm{ }}m/{s^2}} \right) \times {{\left( {1.00{\rm{ }}m} \right)}^2}}}{{\left( {9 \times {{10}^9}{\rm{ }}N \times {m^2}/{C^2}} \right) \times {{\left( {1.6 \times {{10}^{ - 19}}{\rm{ }}C} \right)}^2}}}} \\ &= \sqrt {\frac{{\left( {5.00{\rm{ }}g} \right) \times \left( {\frac{{1{\rm{ }}kg}}{{{{10}^3}{\rm{ }}g}}} \right) \times \left( {9.8{\rm{ }}m/{s^2}} \right) \times {{\left( {1.00{\rm{ }}m} \right)}^2}}}{{\left( {9 \times {{10}^9}{\rm{ }}N \times {m^2}/{C^2}} \right) \times {{\left( {1.6 \times {{10}^{ - 19}}{\rm{ }}C} \right)}^2}}}} \\ &= 1.458 \times {10^{13}}\end{aligned}\)

Fraction of electrons removed

The fraction of electron removed is,

\(\begin{aligned} \frac{n}{{{N_e}}} &= \frac{{1.458 \times {{10}^{13}}}}{{1.433 \times {{10}^{24}}}}\\ &= 1.02 \times {10^{ - 11}}\end{aligned}\)

Hence, the fraction of electrons removed from the nickel atom and placed \(1.00{\rm{ m}}\) above it to support the weight the of five cent coin is \({\rm{1}}{\rm{.02}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 11}}}}\).