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A beam of some kind of particle of velocity $3 \times 10^{7} m / s$ is scattered by an element (Z=50) foil. The specific charge of this particle (charge/mass) is found to be $10^{y}$ (Coulombs/Kg) if the distance of closest approach is $1.6 \times 10^{- 14} m .$ Find the value of y.
[Given $K = 9 \times 10^{9} i n M K S$ ]

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answer is 8.

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Detailed Solution

$\frac{1}{2} m V^{2} = \frac{K q_{1} q_{2}}{r}$

rearranging

$\frac{q_{2}}{m} = \frac{r V^{2}}{2 K q_{1}}$

$\frac{q_{2}}{m} = \frac{1.6 \times 10^{- 14} \times {(3 \times 10^{7})}^{2}}{2 \times 50 \times 1.6 \times 10^{- 19} \times 9 \times 10^{9}} = 10^{8}$

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