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Travka [436]
4 years ago
15

Item 4 part a if a proton and an electron are released when they are 4.50×10−10 m apart (typical atomic distances), find the ini

tial acceleration of each of them.
Chemistry
1 answer:
Wewaii [24]4 years ago
6 0

Acceleration is defined as the rate of change of velocity with respect to time.

Formulas of force are given by:

F=ma   (1)

where,

F = force

m = mass

a = acceleration

F=\frac{kq_{1}q_{2}}{r^{2}}   (2)

where,

F = force

k = coulomb constant (9\times 10^{9}Nm^{2}C^{-2})

r = distance between the charged particles

q_{1} and q_{2} are the signed magnitude of charges

Use the formula (2) for calculating the value of force, we get:

Substitute the value of q_{1} and q_{2}, k and r to find the value of force.

F= \frac{9\times 10^{9} Nm^{2}C^{-2}\times (1.6\times 10^{-19} C)^{2}}{(4.50\times 10^{-10})^{2}}

= 1.1378\times 10^{-9} N

Now, put the above value force in formula (1) to identify the initial acceleration.

1.1378\times 10^{-9} N=ma   (1)   (mass of electron  =9.1 \times 10^{-31} kg)

acceleration of electron = \frac{1.1378\times 10^{-9} N}{9.1\times 10^{-31}kg}

= 1.25\times 10^{21}N/kg

And,

acceleration of proton =\frac{1.1378\times 10^{-9} N}{mass of proton}

mass of proton =1.67 \times  10^{-27}

Thus,

acceleration of proton = \frac{1.1378\times 10^{-9} N}{1.67 \times  10^{-27} kg}

=6.81 \times  10^{17} N/kg

Now, initial acceleration of electron and proton is  1.25\times 10^{21}m/s^{2} and 6.81 \times  10^{17} m/s^{2} as 1 N = kgm/s^{2}.

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