Math  /  Algebra

Question1. Calculate the electrostatic potential energy between two sodium ions ( Na+\mathrm{Na}^{+}) separated by 0.5 nm . Given that the atomic number of sodium is Z=11,14πε0=\mathrm{Z}=11, \frac{1}{4 \pi \varepsilon_{0}}= 9x109Nm2/C9 x 10^{9} \mathrm{Nm}^{2} / \mathrm{C} and the charge on an electron is 1.6×1019C1.6 \times 10^{-19} \mathrm{C}.
2. Two chloride ions (Cl)\left(\mathrm{Cl}^{-}\right)are placed 0.4 nm apart. Using the atomic number of chlorine ( Z=17Z=17 ), calculate the electrostatic potential energy of this system. Assume that the atomic number of sodium is Z=11,14πε0=\mathrm{Z}=11, \frac{1}{4 \pi \varepsilon_{0}}= 9×109Nm2/C9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C} and the charge on an electron is 1.6×1019C1.6 \times 10^{-19} \mathrm{C}.

Studdy Solution
Calculate the electrostatic potential energy between two chloride ions.
2.1 Convert the separation distance from nanometers to meters: $ 0.4 \, \text{nm} = 0.4 \times 10^{-9} \, \text{m} \]
2.2 Use the formula for electrostatic potential energy: $ U = \frac{k \cdot q_1 \cdot q_2}{r} \] where \( q_1 = q_2 = -1 \times 1.6 \times 10^{-19} \, \text{C} \), \( r = 0.4 \times 10^{-9} \, \text{m} \), and \( k = 9 \times 10^{9} \, \text{Nm}^2/\text{C}^2 \).
2.3 Substitute the values into the formula: $ U = \frac{9 \times 10^{9} \cdot (1.6 \times 10^{-19})^2}{0.4 \times 10^{-9}} \]
2.4 Calculate the potential energy: U = \frac{9 \times 10^{9} \cdot 2.56 \times 10^{-38}}{0.4 \times 10^{-9}} \] U = \frac{23.04 \times 10^{-29}}{0.4 \times 10^{-9}} \] $ U = 57.6 \times 10^{-20} \, \text{J} \]

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