An Alfa particles placed at a certain distance from an electron experience an attractive force of 6×10 -11 N .than what is the distance between the particle ?
Answers
Answer:
because the attractive force is 6×10-11 N the distance should be 60-11=49N
The distance between the alpha particle and the electron can be calculated using coulombs law.
1)coulombs law is F=KQ×q/r^2
2)since the force of attraction is given as 6×10^-11 N we can easily calculate the distance between them by using the above formula.
3)Q=charge of alpha particle,. q= charge of an electron while r=distance between them.
4)charge Q of an alpha particle is 3.2×10^-19c
5)charge q on an electron is 1.602×10^-19c
6)The value of constant k is 9×10^9 NM^2
7 ) APPLYING all these quantities by using coulombs law we will get the distance.
8). 6×10^-11=3.2×10^-19 × -1.602×10^-19c×9×10^9/r^2
9)doing the calculations the distance between the alpha particle and electron is 2.77 ×10^-18m.