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?
(N/A) In the alpha-particle scattering experiment,the scattering of $\alpha$-particles is caused by the electrostatic repulsion between the positively charged $\alpha$-particle and the positively charged nucleus of the target atom.
For significant scattering (especially large-angle scattering),the mass of the target nucleus must be significantly greater than the mass of the incident $\alpha$-particle.
The mass of a gold nucleus $(A \approx 197)$ is much larger than the mass of an $\alpha$-particle $(A = 4)$.
However,the mass of a hydrogen nucleus (a proton,$A = 1$) is much smaller than the mass of an $\alpha$-particle.
If solid hydrogen is used as the target,the $\alpha$-particle would not be deflected at large angles because the target nucleus is too light to cause a significant recoil or reflection of the heavier $\alpha$-particle.
Therefore,we would not observe the characteristic large-angle scattering required to determine the size of the nucleus.
For significant scattering (especially large-angle scattering),the mass of the target nucleus must be significantly greater than the mass of the incident $\alpha$-particle.
The mass of a gold nucleus $(A \approx 197)$ is much larger than the mass of an $\alpha$-particle $(A = 4)$.
However,the mass of a hydrogen nucleus (a proton,$A = 1$) is much smaller than the mass of an $\alpha$-particle.
If solid hydrogen is used as the target,the $\alpha$-particle would not be deflected at large angles because the target nucleus is too light to cause a significant recoil or reflection of the heavier $\alpha$-particle.
Therefore,we would not observe the characteristic large-angle scattering required to determine the size of the nucleus.
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View SolutionGiven below are two statements:
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