Lithium is the atom which has three protons in its nucleus and three electrons circling the nucleus.(it usuallly has 3 nuetrons as well,but they r not important in this problem.)Two of the electrons circle the nucleus at a distance 1.8* 10^-11 m (app.)The third electron is farther away at about 1.4 * 10^-10 m from the nucleus and is called the valence electron.

a)Treating the nucleus as a point of charge of value +3e find the force on the valence electron.
b)The electrical force is what supplies the centripetal acceleration (v^2/r) necessary for the electrons circular motion.With what speed does the valence electron circle the nucleus?
c)Find the ionisation energy i.e the energy needed by an electron to completely escape its nuclues, for the lithium atom's valence electron.The ionisation energy is the negative of the electrons total energy E=1/2mv^2 + PE. the PE is that of an electron starting from infinity and being brought to the distance given above from the point charge nucleus.

On a), don't forget that the force on the valence electron includes the attractive force of the nuclues, and the repulsive forces from the inner electrons. So, for a gross simplification, assume a net charge of +1e at the nucleus. Yes, this is a gross simplifiation, but assuming the electrons are circling the nucleus is just as big assumption. Use coulombs law to find the force between the nucleus (+e), and the valence electron (-e).

b) F=mv^2/r solve for v.
c) to find PE, assuming you know calculus, PE=int Force(x)*dx from inf to r. If you don't know calculus, it should work out to ke^2/r .
b)

hey Mr. Bob

I was just wondering why did u take the charge of nucleus +e instead of 3e which is unclear to me..also we gotto find the total force on valence electron ie from both nucleus and the two electrons..am i correct....if not can u help me out..

To solve these problems, we will use the principles of electrostatics and circular motion. Let's go step by step:

a) Force on the valence electron:
We know that the force between two charges can be calculated using Coulomb's law: F = k * (q1 * q2) / r^2, where F is the force, k is the electrostatic constant (approximately 9 × 10^9 N m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.

In this case, the valence electron has a charge of -e (where e is the elementary charge), and the nucleus has a charge of +3e. The distance between them is 1.4 * 10^-10 m. Plugging these values into the formula:

F = (9 × 10^9 N m^2/C^2) * ((-e) * (+3e)) / (1.4 * 10^-10 m)^2

Simplifying:

F = (27 × 9 × 10^9 N m^2/C^2 * e^2) / (1.96 × 10^-20 m^2)

F = 27 * (9 / 1.96) * (10^9 / 10^-20) * e^2 N

F ≈ 123.2 * 10^29 * e^2 N
≈ 1.232 × 10^31 e^2 N

Therefore, the force on the valence electron is approximately 1.232 × 10^31 times the charge of an electron.

b) Speed of the valence electron:
The electric force supplies the centripetal acceleration required for the circular motion of the electron, given by the equation: F = m * (v^2 / r), where m is the mass of the electron, v is its velocity, and r is the distance from the nucleus.

Using this equation, we can equate the electric force (calculated in part a) to the centripetal force:

F = m * (v^2 / r)

1.232 × 10^31 e^2 N = (9.11 × 10^-31 kg) * (v^2 / (1.4 × 10^-10 m))

Simplifying:

v^2 = (1.232 × 10^31 e^2 N * 1.4 × 10^-10 m) / (9.11 × 10^-31 kg)
v^2 ≈ 1.807 × 10^20 e^2 N kg/m

Taking the square root:

v ≈ √(1.807 × 10^20 e^2 N kg/m)

Therefore, the speed of the valence electron is approximately the square root of (1.807 × 10^20 times the charge of an electron) N kg/m.

c) Ionization energy:
The ionization energy is the energy required for the electron to escape the influence of the nucleus and move infinitely away. It can be obtained by calculating the total energy of the electron at its current position (given by its kinetic energy - 1/2 * m * v^2) and subtracting the potential energy associated with moving the electron from infinity to its current position.

Total energy (E) = Kinetic energy (KE) + Potential energy (PE)

The kinetic energy can be calculated using the equation: KE = 1/2 * m * v^2, where m is the mass of the electron and v is its velocity.

The potential energy (PE) is given by the equation: PE = (k * (q1 * q2)) / r, where k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between them.

PE = (9 × 10^9 N m^2/C^2) * ((-e) * (+3e)) / (1.4 × 10^-10 m)

Simplifying:

PE = (27 × 9 × 10^9 N m^2/C^2 * e^2) / (1.4 × 10^-10 m)
≈ 175.5 e^2 J

Now, we can calculate the total energy:

E = KE + PE
E ≈ (1/2 * (9.11 × 10^-31 kg) * v^2) + 175.5 e^2 J

Substituting the value of v^2 calculated in part b:

E ≈ (1/2 * (9.11 × 10^-31 kg) * (1.807 × 10^20 e^2 N kg/m))^2 + 175.5 e^2 J

Therefore, the ionization energy of the valence electron of a lithium atom is approximately the kinetic energy (given by the square of 1.807 × 10^20 times the charge of an electron) J plus 175.5 times the charge of an electron J.