caitlinmizell31
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Consider the electron in the Li2+ cation, initially at rest infinitely away from the nucleus. What will be the velocity when the electron comes to distance R from the nucleus?

Sagot :

Newton's second law and Coulomb's law allow us to find the answer for the electron velocity for the lithium ion at a distance R

         v = [tex]\frac{17.2}{\sqrt{R} }[/tex]

for the ionic radius this velocity is v = 2.2 10⁶ m / s

Newton's second law gives the relationship between force, mass and the acceleration of bodies

            F = ma

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration of the body.

Coulomb's Law gives the value of the electric force that is proportional to the charges and inversely proportional to the square of the distance

         F = [tex]k \frac{q_1q_2}{r^2}[/tex]

Where k is the Coulomb constant (k = 8.99 10⁹ [tex]\frac{N \ m^2}{C^2}[/tex]), q are the charges and r the distance  

When the electron reaches the lithium ion it has a circular motion, therefore the acceleration is centripetal

         a = [tex]\frac{v^2}{r}[/tex]

Where v is the speed (modulus of the velocity) and r the radius of the orbit

we substitute

      [tex]k \frac{q_1q_2}{r^2} = m \frac{v^2}{r}[/tex]

      [tex]v^2 = k \frac{q_1q_2 }{ m \ r}[/tex]k q1q2 / m r = v²

      v = [tex]\sqrt{k \frac{q_1q_2}{m} } \ \frac{1}{\sqrt{r} }[/tex]

In the periodic table we find that the atomic number of the lithium atom is 3 therefore it has 3 positive protons in its nucleus, therefore the charges are

  • The charge of ion Li

       q₁ = + 3 e

  • The charge of the electron

          q₂ = - e

  • The distance is indicated r = R
  • The mass of the electron is m = 9.1  10⁻³¹ kg

Let's substitute

      v = [tex]\sqrt{ \frac{8.99 \ 10^9 \ 3(1.6 \ 10^{-19} )^2 } {9.1 \ 10^{-31}} } \ \frac{1}{\sqrt{R} }[/tex]

      v = [tex]\sqrt{296.374} \ \frac{1}{\sqrt{R} }[/tex]

      v = [tex]\frac{17.2}{\sqrt{R} }[/tex]

We can see that the velocity decreases with the inverse of the square root of the distance from the nucleus, that is, the further away the electron is, its velocity is lower.

To carry out an explicit calculation, let's find the velocity for the ionic radius of Lithium, which is found in the periodic table.

            R = 6 10⁻² nm ([tex]\frac{1 m}{ 10^9 nm}[/tex]) = 6 10⁻¹¹ m

           v = [tex]\frac{17.2}{\sqrt{6 \ 10^{-11}} }[/tex]

           v = 2.2 10⁶ m / s

In conclusion using Newton's second law and Coulomb's law you can find the answer for the electron velocity for the lithium ion at a distance R

         v = [tex]\frac{17.2}{\sqrt{R} }[/tex]  

Learn more about coulomb's law here:

https://brainly.com/question/15077320

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