energy conservation
1/2 mev2 = kZe2/ro
v = √(2 kZe2/mero)
k = 9 * 10^9
Z = 29 for Cu
me = mass of electron
so v = ... aar calc.. chahiye
With what velocity should alpha particle travel towards the nucleus of a copper atom so as to arrive at a distance of 10-13 metre from the nucleus of the copper atom.
when the \alpha particle is placed at a distance of 10-13m to the Cu atom, there will be repulsion between the two nuclei
potential energy due to repulsion =2ze2/4∩kr (k = epsilon)
this shud be equal to K.e of \alpha particle approaching the nucleus =1/2mv2
mv2/2 = 2ze2/4∩kr
v2=4ze2/4∩kmr
solvinf we get
v=6.32x106
energy conservation
1/2 mev2 = kZe2/ro
v = √(2 kZe2/mero)
k = 9 * 10^9
Z = 29 for Cu
me = mass of electron
so v = ... aar calc.. chahiye