Question

A projectile of mass m ,change z ,initial speed u and impact parameter b is scattered by a heavy nucleus of charge Z as shown in figure .Use angular momentum and energy conservation to obtain a formula connecting the minimum distance of the projectile from the nucleus to these parameters .Show that for b=0
,the distance s reduces to the distance of the closest approach r0 ,given by the formula.

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$\tt \: r _{0} = \dfrac{1}{4 \pi \varepsilon _{0}}. \dfrac{z \: z ^{ '} {e}^{2} }{ \frac{1}{2} m {u}^{2} }\ \textless \ br /\ \textgreater \ ​$
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Answer 1

Answer:

: the difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other.

Answer 2

$\bigstar$Solution:-

•In the given figure it shows the trajectory of the projectile having mass m ,velocity u and charge Z'e when scattered by a nucleus having charge Ze ,When it is far off from the nucleus ,it has only K.E ,which is given by;

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$\sf \bullet \: \: E_k = \dfrac{1}{2} m {u}^{2}$

When it scattered from the nucleus at a distance s,then it has both K.E and P.E. These are given by

$\sf \bullet \: \: E_k \: ' = \dfrac{1}{2} m {u'}^{2}$

$\sf \: and \: \bullet \: E_p \: ' = \dfrac{1}{4 \pi \varepsilon_0} . \dfrac{(Z e)\:(Z'e) }{s}$

$\: \: \: \: \: \: \: \sf \implies \dfrac{1}{4 \pi \varepsilon_0} . \dfrac{Z \: Z' {e}^{2} }{s}$

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•Here, u' is the speed of the electron ,when it is at distance s from the nucleus.

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$\sf \therefore\: E_k \: ' +E_p \: ' = \dfrac{1}{2} m {u'}^{2} + \dfrac{1}{4 \pi \varepsilon_0} . \dfrac{Z \: Z' {e}^{2} }{s}$

According to law of conservation of energy

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$\bullet \sf \dfrac{1}{2} m {u'}^{2} + \dfrac{1}{4 \pi \varepsilon_0} . \dfrac{Z \: Z' {e}^{2} }{s} = \dfrac{1}{2} m {u}^{2} \: \: ....(1)$

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•Angular momentum of the projectile, when it is far off from the nucleus

=mu ×h

•The projectile ,when it is at minimum distance from the nucleus

=mu'×s

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According to law of conservation of angular momentum

•mu×b=mu'×s

$\sf or \: u ' = \dfrac{ub}{s}$

In the equation (1) ,Substituting for u' ,we have

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$\implies\sf \dfrac{1}{2} m { \bigg(\dfrac{ub}{s} \bigg)}^{2} + \dfrac{1}{4 \pi \varepsilon_0} . \dfrac{Z \: Z' {e}^{2} }{s} = \dfrac{1}{2} m {u}^{2}$

$\implies\sf \dfrac{1}{2} m {u}^{2} {b}^{2} + \dfrac{1}{4 \pi \varepsilon_0} . {Z \: Z' {e}^{2} }{s} = \dfrac{1}{2} m {u}^{2} \: \: ....(2)$

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•When b=0 ,the projectile will go up to the distance of closest approach r0.Therefore ,in the equation (2) ,setting this condition i.e ,when b=0 ,s =r0, we have

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$\implies\sf \dfrac{1}{2} m {u}^{2} {0}^{2} + \dfrac{1}{4 \pi \varepsilon_0} . {Z \: Z' {e}^{2} }{r_0} = \dfrac{1}{2} m {u}^{2} {(r_0)}^{2}$

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$\sf \implies \dfrac{1}{4 \pi \varepsilon_0} . {Z \: Z' {e}^{2} }{r_0} = \dfrac{1}{2} m {u}^{2} {(r_0)}^{2}$

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$\boxed{\boxed{ \implies \bf r_{0} = \dfrac{1}{4 \pi \varepsilon _{0}} \times \dfrac{Z \: Z '\: e_2}{ \frac{1}{2}m {u}^{2} }}}$