How Do You Calculate the Electron's Energy in the Bohr Model?

[Soaring Crane]

Homework Statement

In the Bohr model of the hydrogen atom, a single electron revolves around a single proton in a circle of radius r. Assume that the proton remains at rest.

a. By equating the electric force to the electron mass times its acceleration, derive an expression for the electron's speed. Express your answer in terms of electron`s charge e, its mass m and orbit radius r.
b. Obtain an expression for the electron's kinetic energy.
c. Obtain an expression for the total energy.
d. Calculate the total energy using 5.29*10^-11 m. Give your answer in joules.
e. Give the answer of part (d) in eV.

Homework Equations

See below.

The Attempt at a Solution

a. F = m(v^2/r)

(k*Q^2)/(r^2) = m*(v^2/r)

k*Q^2*r = m*v^2*r^2, where Q = e = 1.60*10^-19 C

v = sqrt[(k*e^2)/(m*r)]??


b. KE = (m*v^2)/2 = 0.5*[(e^2*k)/(r)] = (e^2)/(8*pi*episilon_0*r) ??

c. E_total = KE + PE = [(e^2)/(8*pi*episilon_0*r)] + [(-e^2)/(4*pi*epsilon_0*r)]



d. E_ total = (e^2)/(8*pi*episilon_0*r) + (-e^2)/(4*pi*epsilon_0*r)

= [(e^2)/(4*pi*epsilon_0*r)]*[0.5 – 1]

= (-0.5)*[1.6*10^-19)^2]/[4*pi*epsilon_0*(5.29*10^-11 m)] = -2.18*10^-18 J ?



E. (-2.18*10^-18 J)/(1.602*10^-19 J/eV) = -13.6 eV?

Thanks.

 

[Hootenanny]

a. F = m(v^2/r)

(k*Q^2)/(r^2) = m*(v^2/r)

k*Q^2*r = m*v^2*r^2, where Q = e = 1.60*10^-19 C

v = sqrt[(k*e^2)/(m*r)]??

Correct.

b. KE = (m*v^2)/2 = 0.5*[(e^2*k)/(r)] = (e^2)/(8*pi*episilon_0*r) ??

Correct.

c. E_total = KE + PE = [(e^2)/(8*pi*episilon_0*r)] + [(-e^2)/(4*pi*epsilon_0*r)]

Correct, although the result can be somewhat simplified;

$$E_{T} =- \frac{e^2}{8\pi\epsilon_0r}$$

d. E_ total = (e^2)/(8*pi*episilon_0*r) + (-e^2)/(4*pi*epsilon_0*r)
= (-0.5)*[1.6*10^-19)^2]/[4*pi*epsilon_0*(5.29*10^-11 m)] = -2.18*10^-18 J ?

I'm not checking your math for you, but assuming you can use a calculator your answer will be correct

E. (-2.18*10^-18 J)/(1.602*10^-19 J/eV) = -13.6 eV?

Assuming your answer to (d) was correct, then this answer is also correct and it tallies with the ground state of hydrogen .

So full marks