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1900's Physics and Wave

The distance ...

Question

The distance of the closest approach of an alpha particle fired at a nucleus with kinetic energy $K_{1}$ is $r_{o}$ . The distance of the closest approach when the α – particle is fired at the same nucleus with kinetic energy $2 K_{1}$ will be.

\frac{r_{o}}{2}

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\frac{r_{o}}{4}

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2 r_{o}

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4 r_{o}

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Solution

The correct option is A $\frac{r_{o}}{2}$
At a distance of closest approach, kinetic energy of α particle = Potential energy of system.
$∴ \frac{K q_{1} q_{2}}{r} = \frac{1}{2} m v^{2}$
Initially kinetic energy,
$K_{1} = \frac{K q_{1} q_{2}}{r_{0}}$ .....(i)
When $K_{2} = 2 K_{1}$
$2 K_{1} = \frac{K q_{1} q_{2}}{r^{'}}$ ....(ii)
From the above equation we get,
$r^{'} = \frac{r_{o}}{2}$

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1900's Physics and Wave

Standard XIII Chemistry

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The distance of the closest approach of an alpha particle fired at a nucleus with kinetic energy K1 is ro. The distance of the closest approach when the α – particle is fired at the same nucleus with kinetic energy 2K1 will be.

The correct option is A ro2At a distance of closest approach, kinetic energy of α particle = Potential energy of system. ∴Kq1q2r=12mv2 Initially kinetic energy, K1=Kq1q2r0 .....(i) When K2=2K1 2K1=Kq1q2r′ ....(ii) From the above equation we get, r′=ro2

The distance of the closest approach of an alpha particle fired at a nucleus with kinetic energy $K_{1}$ is $r_{o}$ . The distance of the closest approach when the α – particle is fired at the same nucleus with kinetic energy $2 K_{1}$ will be.