If a proton and an electron are releaseed whan nay are 750ti

If a proton and an electron are releaseed whan nay are 7.50times10^-10w m apart (typical atom* distarnces find the initial accelrration of each of them.

Solution

Charge of proton

q₁=1.6*10^-19C

Charge of electron

q₂=-1.6*10^-19C

From coloumbs law ,Force of attraction between the 2 charges

F=K|Q₁||Q₂|/d² =(9*10⁹)(1.6*10^-19)/(7.5*10^-10)²

F=4.1*10^-10 N

a)

Mass of proton

M_p=1.67*10^-27 Kg

so initial acceleration of proton

a=F/m =4.1*10^-10/(1.67*10^-27)

a =2.45*10¹⁷ m/s²

b)

Mass of electron

M_e=9.11*10^-31 Kg

so initial acceleration of proton

a=F/m =4.1*10^-10/(9.11*10^-31)

a =4.5*10²⁰ m/s²