Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below $14\; K$.) What results do you expect?
Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below $14\; K$.) What results do you expect?
In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen is less than the mass of incident $\alpha$ - particles Thus, the mass of the scattering particle is more than the target nucleus (hydrogen). As a result, the $\alpha$ particles would not bounce back if solid hydrogen is used in the aparticle scattering experiment and so we cannot determine size of the hydrogen nucleus.
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In an alpha particle scattering experiment distance of closest approach for the $\alpha$ particle is $4.5 \times 10^{-14} \mathrm{~m}$. If target nucleus has atomic number $80$ , then maximum velocity of $\alpha$-particle is . . . . .. $\times 10^5$ $\mathrm{m} / \mathrm{s}$ approximately.
$\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{SI}\right.$ unit, mass of $\alpha$ particle $=$ $\left.6.72 \times 10^{-27} \mathrm{~kg}\right)$
In an alpha particle scattering experiment distance of closest approach for the $\alpha$ particle is $4.5 \times 10^{-14} \mathrm{~m}$. If target nucleus has atomic number $80$ , then maximum velocity of $\alpha$-particle is . . . . .. $\times 10^5$ $\mathrm{m} / \mathrm{s}$ approximately.
$\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{SI}\right.$ unit, mass of $\alpha$ particle $=$ $\left.6.72 \times 10^{-27} \mathrm{~kg}\right)$
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