If a proton and an electron are released when they are 2.0 \times 10^{- 10}m

enfurezca3x 2021-11-23 Answered
If a proton and an electron are released when they are \(\displaystyle{2.0}\times{10}^{{-{10}}}{m}\) apart (a typical atomic distance), find the initial acceleration of each particle.

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Expert Answer

Louis Gregory
Answered 2021-11-24 Author has 8862 answers

Step 1
Given
We are given the distance between a proton and an electron \(\displaystyle{r}={2.0}\times{10}^{{-{10}}}{m}\). We are asked to calculate initial acceleration of each particle.
Solution
The two particles have an opposite charge, therefore each one exerts a force F on the other particle. This force could be calculated by the Coulomb's law as in equalion 21.2
\(F=\frac{1}{4\pi\epsilon_{0}}\frac{\mid q_{p}q_{e}\mid}{r^{2}}\)
Where \(\displaystyle\epsilon_{{{0}}}\) is the electric constant and the term \(\displaystyle{\frac{{{1}}}{{{4}\pi\epsilon_{{{0}}}}}}\) equals \(\displaystyle{9.0}\times{10}^{{{9}}}{N}\cdot\frac{{m}^{{{2}}}}{{C}^{{{2}}}}\). \(\displaystyle{q}_{{{p}}}\) is the charge of proton and equals \(\displaystyle{1.60}\times{10}^{{-{19}}}{C}\) and \(\displaystyle{q}_{{{e}}}\), is the charge of electron and equals \(\displaystyle{1.60}\times{10}^{{-{19}}}{C}\) (See Appendix F)
To obtain the initial acceleration a of the two particles we should get the force that exerted on the proton and the electron. Where the acceleration a is given by Newton's second law in the form.
\(\displaystyle{a}={\frac{{{F}}}{{{m}}}}\)
Where m is the mass of the particle, and for proton \(\displaystyle{m}_{{{p}}}={1.67}\times{10}^{{-{19}}}{k}{g}\) while for electron \(\displaystyle{m}_{{{e}}}={9.11}\times{10}^{{-{31}}}{k}{g}\) (See Appendix F)
Now we can plug our values tor \(\displaystyle{g}_{{{p}}}\), \(\displaystyle{g}_{{{e}}}\) and \(\displaystyle\tau\) into equation (1) to get F
\(F=\frac{1}{4\pi\epsilon_{0}}\frac{\mid q_{p}q_{e}\mid}{r^{2}}\)
\(\displaystyle={\left({9.0}\times{10}^{{{9}}}{N}\cdot\frac{{m}^{{{2}}}}{{C}^{{{2}}}}\right)}{\frac{{{\left({1.60}\times{10}^{{-{19}}}{C}\right)}^{{{2}}}}}{{{\left({2.0}\times{10}^{{-{10}}}{m}\right)}^{{{2}}}}}}\)
\(\displaystyle={5.76}\times{10}^{{-{9}}}{N}\)
Step 2
Now, use equation (2) to get the acceleration a for each particle using the value of F.
For proton: We can plug our values for F and \(\displaystyle{m}_{{{p}}}\) into equation (2) to get the initial acceleration a of the proton
\(\displaystyle{a}_{{{p}}}={\frac{{{F}}}{{{m}_{{{p}}}}}}={\frac{{{5.76}\times{10}^{{-{9}}}{N}}}{{{1.673}\times{10}^{{-{27}}}{k}{g}}}}={3.4}\times{10}^{{{18}}}\frac{{m}}{{s}^{{{2}}}}\)
For electron Again plug our values for F and \(\displaystyle{m}_{{{n}}}\), into equation (2) to gel the initial acceleration a of the electron
\(\displaystyle{a}_{{{e}}}={\frac{{{F}}}{{{m}_{{{p}}}}}}={\frac{{{5.76}\times{10}^{{-{9}}}{N}}}{{{9.109}\times{10}^{{-{31}}}{k}{g}}}}={6.3}\times{10}^{{{21}}}\frac{{m}}{{s}^{{{2}}}}\)

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Ourst1977
Answered 2021-11-25 Author has 7293 answers

image
Proton
\(\displaystyle{m}_{{{p}}}={1.63}\times{10}^{{-{27}}}{k}{g}\)
\(\displaystyle{q}_{{{p}}}={e}={1.602}\times{10}^{{-{19}}}{C}\)
Electron
\(\displaystyle{m}_{{{p}}}={9.109}\times{10}^{{-{31}}}{k}{g}\)
\(\displaystyle{q}_{{{p}}}=-{e}=-{1.602}\times{10}^{{-{19}}}{C}\)
Since the charges are opposite, the Coulon b fuse is attractive.
\(F=\frac{1}{4\pi\epsilon_{0}}\frac{\mid q_{p}q_{e}\mid}{r^{2}}=\left(8.988\times10^{9}\frac{Nm^{2}}{c^{2}}\right)\frac{\left(1.602\times10^{-19}C\right)^{2}}{\left(2.0\times10^{-10}m\right)^{2}}=5.766709788\times 10^{-9}\)
From Newtois Second Law, \(\displaystyle{F}={m}{a}\Rightarrow{a}={\frac{{{F}}}{{{m}}}}\)
\(\displaystyle\therefore\) acceleration of Proton:
\(\displaystyle{a}_{{{p}}}={\frac{{{F}}}{{{m}_{{{p}}}}}}={\frac{{{5.766709788}\times{10}^{{-{9}}}{N}}}{{{1.673}\times{10}^{{-{27}}}{k}{g}}}}={3.4469}\times{10}^{{{18}}}{\frac{{{m}}}{{{v}^{{{2}}}}}}\)
\(\displaystyle\therefore{a}_{{{p}}}={3.4}\times{10}^{{{18}}}{\frac{{{m}}}{{{v}^{{{2}}}}}}\)
acceleration of electron:
\(\displaystyle{a}_{{{e}}}={\frac{{{F}}}{{{m}_{{{e}}}}}}={\frac{{{5.766709788}\times{10}^{{-{9}}}{N}}}{{{9.109}\times{10}^{{-{31}}}{k}{g}}}}={6.33078}\times{10}^{{{27}}}{\frac{{{m}}}{{{v}^{{{2}}}}}}\)
\(\displaystyle\therefore{a}_{{{p}}}={6.3}\times{10}^{{{18}}}{\frac{{{m}}}{{{v}^{{{2}}}}}}\)

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