A neutral lithium atom has three electrons. As you will discover in the textbook

Question

A neutral lithium atom has three electrons. As you will discover in the textbook, two of these electrons form an Inner core," but the third-the valence electron-orbits at much larger radius. From the valence electron's perspective, it is orbiting a spherical ball of charge having net charge +1e (i.e. the three protons in the nucleus and the two inner-core electrons). The energy required to ionize a lithium atom is 5.14 eV. According to Rutherford's nuclear model of the atom, what is the orbital radius of the valence electron? Express your answer with the appropriate units. According to Rutherford's nuclear model of the atom, what is the speed of the valence electron? Express your answer with the appropriate units.

Explanation / Answer

A) Due to shielding, the total energy of the orbiting valence electron E = ½ m v ² - ( k e ² / r )

the force in circular motion is given by F = m v ² / r

while the electric interaction due to shielding gives a force with magnitude F = k e ² / r ²

so that equality of the two forces gives m v ² / r = k e ² / r ² which reduces to
m v ² = k e ² / r the total energy of the orbiting valence electron is therefore

E = ½ m v ² - ( k e ² / r ) = ½ ( k e ² / r ) – ( k e ² / r )  = - ( k e ² / 2 r )

solving for the orbital radius r, we get r = - k e ² / [ 2 E ] where
E = - 5.14 eV = - ( 5.14 eV ) (1.6 × 10 ¹ J / eV ) = - ( 5.14 e ) J
negative energy, as in the total energy considered earlier, since the valence electron is a bound state, it follows that the orbital radius is r = - k e ² / [ 2 E ] = - k e ² / [ 2 ( -5.14 e ) J ] = k e / [ 2 (5.14 ) J ]
= ( 8.99 × 10 N • m ² ) ( 1.6 × 10 ¹ ) / [ 2 (5.14 ) N • m ]
r = 1.399 × 10 ¹ m

B)   the energy of the electron is E=K+U = 1/2mv^2+U
then v=[2(E-U)/m]^1/2
where E=5.14 eV = 8.224 x 10^-19 J
U= -qV = -ke^2/r = -16.47 x 10^-19 J
then
v=[2(8.224+16.47)x10^-19/9.1x10^-31]^1/2
v=2.33x10^6 m/s

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