# Stuck with physics ?

## In this problem you will examine the forces in the Bohr model of the atom. In the Bohr model of the hydrogen atom, an electron of mass 9.1×10−31kg and charge -1.6×10−19C is considered to be moving at a constant speed of 2.1113x10+6 m s−1 in a circular orbit of radius 5.2335x10-11 m around a proton of mass 1.67×10−27kg and charge +1.6×10−19C . Calculate the centripital force, (m v2)/r, necessary to keep the electron moving in its orbit. Give your answer in Newtons. 2.) Calculate the electrostatic force (magnitude) acting on the electron due to the proton. Give your answer in Newtons.

Jan 26, 2018

Centripetal force: $7.7509 \times {10}^{- 8} N$

Electrostatic force: $8.4005 \times {10}^{- 8} N$

#### Explanation:

The centripetal force is given by:

$F = \frac{m {v}^{2}}{r}$ as stated (where m is the mass of the electron):

We have been told that:

$m = 9.1 \times {10}^{- 31} k g$
$v = 2.1113 \times {10}^{6} m {s}^{- 1}$
$r = 5.2335 \times {10}^{- 11} m$

Simply put these numbers into the expression to find the force:

$F = \frac{m {v}^{2}}{r} = \frac{\left(9.1 \times {10}^{- 31}\right) {\left(2.1113 \times {10}^{6}\right)}^{2}}{5.2335 \times {10}^{- 11}}$

$= 7.7509 \times {10}^{- 8} N$

For the electrostatic force use:

$F = \frac{{Q}_{1} {Q}_{2}}{4 \pi {\epsilon}_{o} {r}^{2}}$

In the case:

The charge of the electron is the same but opposite the charge of the neutron so (we want the magnitude so just take the positive of both anyway):

${Q}_{1} = {Q}_{2} = 1.6 \times {10}^{- 19} C$

$\epsilon$ is the permittivity of free space:

$\epsilon = 8.854 \times {10}^{- 12}$

Finally, $r$ is the radius again:

$r = 5.2335 \times {10}^{- 11} m$

Plug these into the formula to get:

$F = \frac{{Q}_{1} {Q}_{2}}{4 \pi {\epsilon}_{o} {r}^{2}} = \frac{\left(1.6 \times {10}^{- 19}\right) \left(1.6 \times {10}^{- 19}\right)}{4 \pi \left(8.854 \times {10}^{- 12}\right) {\left(5.2335 \times {10}^{- 11}\right)}^{2}}$

$8.4005 \times {10}^{- 8} N$

There are discrepancies in the results (e.g. the two forces are out by about $0.65 \times {10}^{- 8} N$. This may be due to assuming a circular orbit. A modified version of the bohr model assumes an elliptical orbit which means the velocity, radius and force don't stay constant.

Any further discrepancies may also may be down to the fact that relativistic effects have not been accounted for.

Jan 26, 2018

I get centripetal force required, ${F}_{c} = 7.748 \times {10}^{-} 8$N and the electrostatic force, ${F}_{e} = 8.404 \times {10}^{-} 8$N i.e. available force just > centripetal force necessary.

#### Explanation:

My calculations are as follows:

${F}_{c} = \frac{m {v}^{2}}{r} = \frac{9.1 \times {10}^{-} 31 \times {\left(2.1113 \times {10}^{6}\right)}^{2}}{5.2335 \times {10}^{-} 11}$

${F}_{c} = 7.748 \times {10}^{-} 8$N

${F}_{e} = \left(\frac{1}{4 \pi {\epsilon}_{0}}\right) \left(\frac{{q}_{1} {q}_{2}}{r} ^ 2\right)$

${F}_{e} = \frac{1}{4 \pi \times 8.85 \times {10}^{-} 12} \left({\left(1.6 \times {10}^{-} 19\right)}^{2} / {\left(5.2335 \times {10}^{-} 11\right)}^{2}\right)$

${F}_{e} = 8.404 \times {10}^{-} 8$N