A) The acceleration of the electron is
towards the proton
B) The electric field magnitude at the electron's location is 
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:
where:
is the Coulomb's constant
is the charge of the proton
is the charge of the electron
is the distance between the electron and the proton
Substituting,

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:

where
F is the force
is the electron mass
Substituting,

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

where
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:
(proton charge)
(location of the electron with respect to the proton)
Therefore, the magnitude of the electric field is

C)
The electric field at which the air begins to spark is

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

By comparing the two fields, we observe that

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:
brainly.com/question/2757829
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