Question

If a proton and an electron are released when they are 4.00 x 10^-10 m apart, find the initial acceleration of each of them.

Answer 1

(A) Acceleration of electron
`a=(1.44* 10^(-9))/(9.11* 10^(-31))=0.158* 10^(22)m/sec^2`

(b) Acceleration of proton
`a=(1.44* 10^(-9))/(1.67* 10^(-27))=0.8622* 10^(18)m/sec^2`

Step-by-step explanation:

We have given distance between proton
`r=4* 10^(-10)m`

Charge on proton and charge on electron
`q=1.6* 10^(-19)`

According to coulombs law force between two charge
`F=(1)/(4\pi \epsilon _0)(q_1q_2)/(r^2)=(Kq_1q_!)/(r^2)=(9* 10^9* 1.6* 10^(-19)* 1.6* 10^(-19))/((4* 10^(-10))^2)=1.44* 10^(-9)N`

Mass of electron
`m=9.11* 10^(-31)kg`

So acceleration of electron
`a=(1.44* 10^(-9))/(9.11* 10^(-31))=0.158* 10^(22)m/sec^2`

Mass of proton
`m=1.67* 10^(-27)kg`

So acceleration of proton
`a=(1.44* 10^(-9))/(1.67* 10^(-27))=0.8622* 10^(18)m/sec^2`

Answer 2

For proton

`a=8.8* 10^(17)\ m/s^2`

For electron

`a=1.5* 10^(21)\ m/s^2`

Step-by-step explanation:

We know that

The mass of electron

`m=9.1* 10^(-31)\ kg`

The mass of proton

`m=1.67* 10^(-27)\ kg`

Charge on electron and proton

q₁=q₂=q

`q=1.6* 10^(-19)\ C`

Electrostatics force

`F=K(q_1q_2)/(r^2)`

Now by putting the values

`F=9* 10^9* (1.6* 10^(-19)* 1.6* 10^(-19))/((4* 10^(-10))^2)`

`F=1.44* 10^(-9)\ N`

For proton

F = m a

a =F/m

`a=(1.4* 10^(-9))/(1.67* 10^(-27))\ m/s^2`

`a=8.8* 10^(17)\ m/s^2`

For electron

`a=(1.4* 10^(-9))/(9.1* 10^(-31))\ m/s^2`

`a=1.5* 10^(21)\ m/s^2`