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

Answer:

For proton

$a=8.8\times 10^{17}\ m/s^2$

For electron

$a=1.5\times 10^{21}\ m/s^2$

Explanation:

We know that

The mass of electron

$m=9.1\times 10^{-31}\ kg$

The mass of proton

$m=1.67\times 10^{-27}\ kg$

Charge on electron and proton

q₁=q₂=q

$q=1.6\times 10^{-19}\ C$

Electrostatics force

$F=K\dfrac{q_1q_2}{r^2}$

Now by putting the values

$F=9\times 10^9\times \dfrac{1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(4\times 10^{-10})^2}$

$F=1.44\times 10^{-9}\ N$

For proton

F = m a

a =F/m

$a=\dfrac{1.4\times 10^{-9}}{1.67\times 10^{-27}}\ m/s^2$

$a=8.8\times 10^{17}\ m/s^2$

For electron

$a=\dfrac{1.4\times 10^{-9}}{9.1\times 10^{-31}}\ m/s^2$

$a=1.5\times 10^{21}\ m/s^2$

Answer 2

Answer:

(A) Acceleration of electron $a=\frac{1.44\times 10^{-9}}{9.11\times 10^{-31}}=0.158\times 10^{22}m/sec^2$

(b) Acceleration of proton $a=\frac{1.44\times 10^{-9}}{1.67\times 10^{-27}}=0.8622\times 10^{18}m/sec^2$          

Explanation:

We have given distance between proton $r=4\times 10^{-10}m$

Charge on proton and charge on electron $q=1.6\times 10^{-19}$

According to coulombs law force between two charge $F=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}=\frac{Kq_1q_!}{r^2}=\frac{9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(4\times 10^{-10})^2}=1.44\times 10^{-9}N$

Mass of electron $m=9.11\times 10^{-31}kg$

So acceleration of electron $a=\frac{1.44\times 10^{-9}}{9.11\times 10^{-31}}=0.158\times 10^{22}m/sec^2$

Mass of proton $m=1.67\times 10^{-27}kg$

So acceleration of proton  $a=\frac{1.44\times 10^{-9}}{1.67\times 10^{-27}}=0.8622\times 10^{18}m/sec^2$