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If a proton and an electron are released when they are 6.50×10⁻¹⁰ m apart (typical atomic distances), find the initial acceleration of each of them.
a) Acceleration of electron
b) Acceleration of proton

Respuesta :

Khoso123

Answer:

(a) Acceleration of electron= 5.993×10²⁰ m/s²

(b) Acceleration of proton= 3.264×10¹⁷ m/s²

Explanation:

Given Data

distance r= 6.50×10⁻¹⁰ m

Mass of electron Me=9.109×10⁻³¹ kg

Mass of proton Mp=1.673×10⁻²⁷ kg

Charge of electron qe= -e = -1.602×10⁻¹⁹C

Charge of electron qp= e = 1.602×10⁻¹⁹C

To find

(a) Acceleration of electron

(b) Acceleration of proton

Solution

Since the charges are opposite the Coulomb Force is attractive

So

[tex]F=\frac{1}{4(\pi)Eo }\frac{|qp*qe|}{r^{2} }\\   F=(8.988*10^{9}Nm^{2}/C^{2})*\frac{(1.602*10^{-19})^{2}  }{(6.50*10^{-10} )^{2}  } \\F=5.46*10^{-10}N[/tex]

From Newtons Second Law of motion

F=ma

a=F/m

For (a) Acceleration of electron

[tex]a=F/Me\\a=(5.46*10^{-10} )/9.109*10^{-31}\\ a=5.993*10^{20}m/s^{2}[/tex]

For(b) Acceleration of proton

[tex]a=F/Mp\\a=(5.46*10^{-10} )/1.673*10^{-27} \\a=3.264*10^{17}m/s^{2}[/tex]

Acceleration of the body is rate of change of the increasing velocity with respect to the time.  is The acceleration of the electron is [tex]5.992\times10^{20}[/tex] meter per second squared and  acceleration of the proton is [tex]3.265\times10^{17}[/tex] meter per second squared.

Given information-

A proton and an electron are released when they are 6.50×10⁻¹⁰ m apart.

The the distance of proton and electron is 6.50×10⁻¹⁰ m.

Now it is known that,

  • Mass of the electron,

        [tex]m_e=9.109\times10^{-31}[/tex]

  • Mass of the proton,

        [tex]m_p=1.673\times10^{-27}[/tex]

  • Charge of the electron,

        [tex]Q_e=-1.602\times10^{-19}[/tex]

  • Charge of the proton,

        [tex]Q_e=1.602\times10^{-19}[/tex]

Acceleration

Acceleration of the body is rate of change of the increasing velocity with respect to the time. Acceleration of a body is the ratio of the force to the mass of the body.

The coulomb force between the two charges can be given as,

[tex]F=k\dfrac{q_1q_2}{r^2} [/tex]

Here k is the coulombs constant, r is the distance and q is the charge of proton and electron.Put the values,

[tex]F=9\times10^9\times\dfrac{(1.602\times10^{-19})^2}{(6.5\times10^{-10})^2} [/tex]

[tex]F=5.46\times10^{-10}[/tex]

Now it is known that the acceleration of a body is the ratio of the force to the mass of the body.

  • Acceleration of electron

[tex]a_e=\dfrac{F}{m_e} [/tex]

[tex]a_e=\dfrac{5.46\times10^{-10}}{9.109\times10^{-31}} [/tex]

[tex]a_e=5.992\times10^{20}[/tex]

  • Acceleration of Proton

[tex]a_p=\dfrac{F}{m_p} [/tex]

[tex]a_p=\dfrac{5.46\times10^{-10}}{1.673\times10^{-27}} [/tex]

[tex]a_e=3.625\times10^{17}[/tex]

Thus the acceleration of the electron is [tex]5.992\times10^{20}[/tex] meter per second squared and  acceleration of the proton is [tex]3.265\times10^{17}[/tex] meter per second squared.

Learn more about the acceleration here;

https://brainly.com/question/12134554

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