Question

A uranium ion and an iron ion are separated by a distance of =61.10 nm. The uranium atom is singly ionized; the iron atom is doubly ionized. Calculate the distance from the uranium atom at which an electron will be in equilibrium. Ignore the gravitational attraction between the particles. = nm An electron sits between a singly ionized uranium ion and a doubly ionized iron ion. The distance from the uranium ion to the electron is designated lowercase r, and the distance between the two ions is designated uppercase R. What is the magnitude of the force on the electron from the uranium ion? magnitude of the force: N

Answer 1

Answer:

Explanation:

Charge on uranium ion = charge of a single electron

= 1.6 x 10⁻¹⁹ C

charge on doubly ionised iron atom = charge of 2 electron

= 2 x 1.6 x 10⁻¹⁹ C = 3.2 x 10⁻¹⁹ C

Let the required distance from uranium ion be d .

force on electron at distance d from uranium ion

= 9 x 10⁹ x 1.6 x 10⁻¹⁹ / r²

force on electron at distance 61.10 x 10⁻⁹ - r from iron  ion

= 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²

For equilibrium ,

9 x 10⁹ x 1.6 x 10⁻¹⁹ / r² = 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²

2 d² = (61.10 x 10⁻⁹ - r )²

1.414 r = 61.10 x 10⁻⁹ - r

2.414 r = 61.10 x 10⁻⁹

r = 25.31 nm .

Answer 2

(a)  The distance from the uranium atom at which an electron will be in equilibrium is $2.04 \times 10^{-8} \ m$

(b) The magnitude of the force on the electron from the uranium ion is $3.46 \times 10^ 6 \ N$

The given parameters:

• distance between the iron and the uranium, d = 61.1 nm
• charge of uranium ion, q₁ = 1.6 x 10⁻¹⁹ C
• charge of doubly ionized atom, q₂ = 2q₁ = 3.2 x 10⁻¹⁹ C

The force on the electron due to uranium ion at distance r is calculated as follows;

$F _1 = \frac{Kq_1^2}{r^2} \\\\F_1 = \frac{9\times 10^9 \times (1.6\times 10^{-19})^2}{r^2} \\\\F_1 = \frac{2.3 \times 10^{-28}}{r^2}$

The force on the electron due to uranium ion at distance less than 61.10 nm.

R = 61.10 nm - r

$F_2 = \frac{9\times 10^9 \times (3.2 \times 10^{-19})^2}{(61.1 \times 10^{-9} \ - \ r)^2} \\\\\F_2 = \frac{9.216 \times 10^{-28}}{(61.1 \times 10^{-9} \ - \ r)^2}$

At equilibrium, the force between the electron and ions will be equal.

$\frac{9.216 \times 10^{-28}}{(61.1 \times 10^{-9} \ - \ r)^2}= \frac{2.3 \times 10^{-28}}{r^2}\\\\\frac{4}{(61.1 \times 10^{-9} \ - \ r)^2} = \frac{1}{r^2} \\\\4r^2 = (61.1 \times 10^{-9} \ - \ r)^2\\\\2^2r^2 = (61.1 \times 10^{-9} \ - \ r)^2\\\\2r = 61.1 \times 10^{-9} \ - \ r\\\\3r = 61.1 \times 10^{-9} \\\\r = \frac{61.1 \times 10^{-9}}{3} \\\\r = 2.04 \times 10^{-8} \ m$

The magnitude of the force on the electron from the uranium ion is calculated as follows;

$F = \frac{kq_1^2}{r^2} \\\\F = \frac{9\times 10^9 \times 1.6\times 10^{-19}}{(2.04 \times 10^{-8})^2} \\\\F= 3.46 \times 10^6 \ N$

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