An - particle experiences the following force due to a nucleus k -0 begin array 11 l 1 F - frac k r - 2- if-r - R- frac - K R - 3vec r - -if- r - R-end array Suppose the particle starts from r - with a kinetic energy just enough to reach r - R- Its kinetic energy at r - R -2 will beA- 0B- 5-8 k - R C- 3-8k-RD- k-R

The correct option is C ( 3/8 ) ( k/R )KE=U_(R) U_ ( R/2 )=k/2R k/8R=3k/8R ...

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Byju's Answer

Standard XII

Physics

Instantaneous Force and Impulse

An α- particl...

Question

An $α -$ particle experiences the following force due to a nucleus $(k > 0)$
$\begin{matrix} \to F = \frac{k}{r^{2}} & i f & r > R = \frac{- K}{R^{3}} \to r & i f & r < R \end{matrix}$

Suppose the particle starts from $r = \infty$ with a kinetic energy just enough to reach r = R. Its kinetic energy at $r = \frac{R}{2}$ will be

\frac{k}{R}

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(\frac{5}{8}) (\frac{k}{R})

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(\frac{3}{8}) (\frac{k}{R})

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Solution

The correct option is C $(\frac{3}{8}) (\frac{k}{R})$
$K E = U_{(R)} - U_{(\frac{R}{2})} = \frac{k}{2 R} - \frac{k}{8 R} = \frac{3 k}{8 R}$

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An α−particle experiences the following force due to a nucleus (k>0) →F=kr2ifr>R=−KR3→rifr<R Suppose the particle starts from r=∞ with a kinetic energy just enough to reach r = R. Its kinetic energy at r=R2 will be

The correct option is C (38)(kR)KE=U(R)−U(R2)=k2R−k8R=3k8R

The correct option is C $(\frac{3}{8}) (\frac{k}{R})$
$K E = U_{(R)} - U_{(\frac{R}{2})} = \frac{k}{2 R} - \frac{k}{8 R} = \frac{3 k}{8 R}$