Question

The charge on an electron is −1.6×10−19 C. The charge on a proton is equal in magnitude, but positive. The electron's mass is 9.11×10−31 kg.
A. In an atom, the electron and the proton and electron are seperated by about 0.5 Angstroms, 5.0×10−11 m. What is the acceleration of an electron of an electron in an atom?

B. What is the electric field magnitude due to the proton at the electron's location?

C. The "dielectric strength" is the electric field magnitude at which a substance begins to spark. Air begins to conduct or spark when the electric field is about 3 million N/C. How does this electric field compare to the magnitude of the electric field in an atom?

1. The electric field in an atom is much less than the dielectric strength of air
2. The electric field in an atom is much greater than the dielectric strength of air
3. The electric field in an atom is within an order of magnitude (a factor of 10) of the dielectric strength of air

Answer 1

A) The acceleration of the electron is $1.0\cdot 10^{23} m/s^2$ towards the proton

B) The electric field magnitude at the electron's location is $5.76\cdot 10^{11} N/C$

C) 2. The electric field in an atom is much greater than the dielectric strength of air

Explanation:

A)

In order to find the acceleration of the electron, we need to find the force acting on it first.

The force on the electron is the electrostatic force between the proton and the electron, which is given by:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1 = 1.6\cdot 10^{-16}C$ is the charge of the proton

$q_2 = -1.6\cdot 10^{-19}C$ is the charge of the electron

$r=5.0\cdot 10^{-11}m$ is the distance between the electron and the proton

Substituting,

$F=(8.99\cdot 10^9) \frac{(1.6\cdot 10^{-19})(-1.6\cdot 10^{-19})}{(5.0\cdot 10^{-11})^2}=-9.2\cdot 10^{-8} N$

where the negative sign indicates the direction of the force is towards the proton (attractive).

Now we can find the acceleration of the electron using Newton's second law:

$a=\frac{F}{m}$

where

F is the force

$m=9.11\cdot 10^{-31} kg$ is the electron mass

Substituting,

$a=\frac{-9.2\cdot 10^{-8}}{9.11\cdot 10^{-31}}=-1.0\cdot 10^{23} m/s^2$

where the negative sign means the direction is towards the proton.

B)

The magnitude of the electric field due to a single-point charge is given by

$E=k\frac{q}{r^2}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge at which the field is calculated

Here we have:

$q=1.6\cdot 10^{-19}C$ (proton charge)

$r=5.0\cdot 10^{-11} m$ (location of the electron with  respect to the proton)

Therefore, the magnitude of the electric field is

$E=(8.99\cdot 10^9)\frac{1.6\cdot 10^{-19}}{(5.0\cdot 10^{-11})^2}=5.76\cdot 10^{11} N/C$

C)

The electric field at which the air begins to spark is

$E_b = 3\cdot 10^6 N/C$

While the electric field in the atom (at the location of the electron), calculated in part B, si

$E=5.76\cdot 10^{11} N/C$

By comparing the two fields, we observe that

$E_b$

which means that the correct statement is

2. The electric field in an atom is much greater than the dielectric strength of air

Learn more about atoms:

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