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What alpha-rays strikea thin gold foil t...

What `alpha`-rays strikea thin gold foil then

A

Most of the `alpha-`rays do not pass through the gold foil

B

Most of the `alpha-` rays get deflected back

C

Most of the`alpha-`rays get deflected through small angles

D

Most of the `alpha-`rays pass through without any deviation.

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To solve the question regarding the behavior of alpha rays striking a thin gold foil, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding Alpha Rays**: Alpha rays are positively charged particles emitted from certain radioactive materials. They consist of two protons and two neutrons, making them relatively heavy compared to other forms of radiation. **Hint**: Remember that alpha rays are positively charged and heavier than beta rays or gamma rays. 2. **Nature of Gold Foil**: Gold foil is made of gold atoms, which have a nucleus containing protons (positive charge) and neutrons. The electrons surrounding the nucleus are negatively charged. **Hint**: Recall that atoms consist of a small, dense nucleus surrounded by a cloud of electrons. 3. **Interaction of Alpha Rays with Gold Foil**: When alpha rays strike the gold foil, most of them pass through the foil without any interaction. This is because the nucleus of the gold atoms is very small compared to the overall size of the atom. **Hint**: Consider the size of the nucleus relative to the entire atom when thinking about the likelihood of interaction. 4. **Deflection of Alpha Rays**: A small fraction of alpha rays may be deflected at small angles when they come close to the nucleus. However, very few are deflected back. This is due to the repulsion between the positively charged alpha particles and the positively charged nucleus. **Hint**: Think about the forces at play between charged particles—like charges repel each other. 5. **Conclusion**: Given the options provided: - Option 1: Most of the alpha rays do not pass through the gold foil. (Incorrect) - Option 2: Most of the alpha rays get deflected back. (Incorrect) - Option 3: Most of the alpha rays get deflected through a small angle. (Incorrect) - Option 4: Most of the alpha rays pass through without any deflection. (Correct) Therefore, the correct answer is **Option 4**: Most of the alpha rays pass through without any deflection. ### Summary: The behavior of alpha rays when they strike a thin gold foil demonstrates that most of them pass through due to the small size of the nucleus compared to the atom, while only a small number experience deflection.
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When alpha - particles are passed through a thin foil, then

(a) Alpha particles having kinetic energy of 1.8 MeV each are incident on a thin gold foil , from a large distance. Applying the principle of conversation of energy, find the closest distance of approach of the alpha particle from the gold nucleus. (Atomic number of gold = 79 ).

The alpha- rays have more penetrating power than gamma- rays.

(A) most of the mass of the atom is concentrated in its nucleus. (R) all alpha particles striking a gold sheet are scattered in different directions.

(A) in the experiment of alpha particle scattering, extremely thin gold foils are preferred over other metals. (R)gold is a ductile material.

Define the distance of closest approach. An alpha -particle of kinetic enegy 'K' is bombarded on a thin gold foil. The distance of the closest approach is 'r' . What will be the distance of closest approach for an alpha -particle of double the kinetic energy ?

In Rutherford's experiment, generally the thin foil of heavy atoms, such as gold, platinum, etc. have been used to be bombarded by the alpha -particles. If the thin foil of light atoms such as aluminium etc. Is used, what difference would be observed form the above results?

Draw a schematic arrangement of the Geiger Marsden experiment. How did the scattering of alpha particles by a thin foil of gold provide an important way to determine an upper limit on the size of nucleus? Explain briefly.

In Rutherford experiments on alpha -ray scattering the number of particles scattered at 90^(0) angle be 28 per minute. Then the number of particles scattered per minute by the same foil, but at 60^(0) is

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