The closest distance of approach of an - - particle travelling with a velocity v towards a stationary nucleus is d- For the closest distance to become d2 towards a stationary nucleus of double the charge- the velocity of projection of the - particle has to be

From work energy theorem, (KE)_α = P.E of Nucleus and α particle 12 mv^2 = 14 πε_0 ( q_n q_αd ) ........(1) For, q^'_n=2q_n d^'=d2 12mv'^2= 14 πε_0 ( q_n^ ...

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Standard XII

Physics

Nucleus: Common Terms

The closest d...

Question

The closest distance of approach of an $α$ - particle travelling with a velocity $v$ towards a stationary nucleus is $d$ . For the closest distance to become $\frac{d}{2}$ towards a stationary nucleus of double the charge, the velocity of projection of the $α$ - particle has to be

\sqrt{2} v

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2 v

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4 v

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\sqrt{6} v

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Solution

The correct option is B $2 v$
From work-energy theorem,

$(K E)_{α}$ = P.E of Nucleus and $α -$ particle

$\frac{1}{2} m v^{2} = \frac{1}{4 π ε_{0}} (\frac{q_{n} q_{α}}{d}) . . . . . . . . (1)$

For, $q_{n}^{^{'}} = 2 q_{n} d^{^{'}} = \frac{d}{2}$

$\frac{1}{2} m v {^{'}}^{2} = \frac{1}{4 π ε_{0}} (\frac{q_{n}^{^{'}} q_{α}}{d^{^{'}}}) . . . . . . . . (2)$

On dividing (2) $\div$ (1)

${(\frac{v^{'}}{v})}^{2} = \frac{q_{n^{'}}}{q_{n}} \times \frac{d}{d^{^{'}}}$

${(\frac{v^{'}}{v})}^{2} = \frac{2 q_{n}}{q_{n}} \times \frac{d}{(\frac{d}{2})} = 4$

$∴ v^{^{'}} = 2 v$

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The closest distance of approach of an α - particle travelling with a velocity v towards a stationary nucleus is d. For the closest distance to become d2 towards a stationary nucleus of double the charge, the velocity of projection of the α- particle has to be

The correct option is B 2vFrom work-energy theorem, (KE)α = P.E of Nucleus and α− particle 12mv2=14πε0(qnqαd) ........(1) For, q′n=2qnd′=d2 12mv′2=14πε0(q′nqαd′) ........(2) On dividing (2) ÷ (1) (v′v)2=qn′qn×dd′ (v′v)2=2qnqn×d(d2)=4 ∴ v′=2v

The closest distance of approach of an $α$ - particle travelling with a velocity $v$ towards a stationary nucleus is $d$ . For the closest distance to become $\frac{d}{2}$ towards a stationary nucleus of double the charge, the velocity of projection of the $α$ - particle has to be