A projectile of mass m, charge Z', initial speed v and impact parameter b is scattered by a heavy nucleus of charge Z. Use angular momentum and energy conservation to obtain a formula connecting the minimum distance (s) of the projectile form the nucleus to these parameters .show that for b=0, s reduces to the closest distance of approach `r_0`.
A projectile of mass m, charge Z', initial speed v and impact parameter b is scattered by a heavy nucleus of charge Z. Use angular momentum and energy conservation to obtain a formula connecting the minimum distance (s) of the projectile form the nucleus to these parameters .show that for b=0, s reduces to the closest distance of approach `r_0`.
Text Solution
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Charge on the nucleus =Ze
Charge on the projectile =Z'e
At infinity, angular momentum of projectile about the nucleus=mvb.
At the point of minimum distance (s) form the nucleus, velocity v' of the projectile is normal to the radius vector (form the centre of nucleus to projectile) , fig.
`:. ` Angular momentum to projectile about the nucleus =mv's
According to the law of conservation of angular momentum
`mv's=mvb or v'=(vb)/s....(i)`
Now, at infinity, K.E. of particle =`1/2 mv^2`
and P.E.=0
At the minimum distance s, K.E of particle =`1/2mv'^2`
P.E of particle `=1/(4pi in_0) ((Ze)(Z'e))/s`
According to the principle of energy conservation,
`1/2mv'^2+1/(4pi in_0) (Z Z'e^2)/s=1/2 mv^2`
Using (1) we get,
`1/2mv^2 (b^2)/(s^2)+1/(4pi in_0) (Z Z'e^2)/s=1/2mv^2`
or `1/2mv^(2)b^(2)+1/(4pi in_0) Z Z'e^2.s=1/2mv^2s^2`
Dividing both sides by `(1/2mv^2)`, we get,
`b^2+1/(2pi in_0) (Z Z'e^2 s)/(mv^2)=s^2`
for a head on collision, b=0
`:. s^2=1/(2pi in_0) (Z Z'e^2 s)/(mv^2)`
or `s=1/(2pi in_0) (Z Z'e^2)/(mv^2)=1/(4pi in_0) ((Ze)(Z'e))/((1/2 mv^2))`
Which is `r_0`, the distance of closest approach.
Charge on the projectile =Z'e
At infinity, angular momentum of projectile about the nucleus=mvb.
At the point of minimum distance (s) form the nucleus, velocity v' of the projectile is normal to the radius vector (form the centre of nucleus to projectile) , fig.
`:. ` Angular momentum to projectile about the nucleus =mv's
According to the law of conservation of angular momentum
`mv's=mvb or v'=(vb)/s....(i)`
Now, at infinity, K.E. of particle =`1/2 mv^2`
and P.E.=0
At the minimum distance s, K.E of particle =`1/2mv'^2`
P.E of particle `=1/(4pi in_0) ((Ze)(Z'e))/s`
According to the principle of energy conservation,
`1/2mv'^2+1/(4pi in_0) (Z Z'e^2)/s=1/2 mv^2`
Using (1) we get,
`1/2mv^2 (b^2)/(s^2)+1/(4pi in_0) (Z Z'e^2)/s=1/2mv^2`
or `1/2mv^(2)b^(2)+1/(4pi in_0) Z Z'e^2.s=1/2mv^2s^2`
Dividing both sides by `(1/2mv^2)`, we get,
`b^2+1/(2pi in_0) (Z Z'e^2 s)/(mv^2)=s^2`
for a head on collision, b=0
`:. s^2=1/(2pi in_0) (Z Z'e^2 s)/(mv^2)`
or `s=1/(2pi in_0) (Z Z'e^2)/(mv^2)=1/(4pi in_0) ((Ze)(Z'e))/((1/2 mv^2))`
Which is `r_0`, the distance of closest approach.
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