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Statement-I : In Rutherford's gold foil ...

Statement-I : In Rutherford's gold foil experiment, very few `alpha`-particles are deflected back.
Because
Statement-II : Nucleus present inside the atom is heavy.

A

Statement-I is true, Statement-II is true , Statement-II is correct explanation for Statement-I.

B

Statement-I is true, Statement-II is true , Statement-II is NOT a correct explanation for statement-I

C

Statement-I is true, Statement-II is false

D

Statement-I is false, Statement-II is true

Text Solution

AI Generated Solution

The correct Answer is:
To analyze the statements given in the question, we can break down the reasoning step by step. ### Step 1: Understand Rutherford's Gold Foil Experiment Rutherford's gold foil experiment was designed to probe the structure of the atom. In this experiment, alpha particles (which are positively charged) were directed at a thin foil of gold. Most of the alpha particles passed through the foil without any deflection, while a small fraction were deflected back. **Hint:** Consider what the experiment was designed to reveal about the atomic structure. ### Step 2: Analyze Statement-I Statement-I states that "In Rutherford's gold foil experiment, very few alpha particles are deflected back." This is indeed true. The majority of the alpha particles passed through the foil without any interaction. **Hint:** Think about the implications of most particles passing through without deflection. ### Step 3: Analyze Statement-II Statement-II claims that "The nucleus present inside the atom is heavy." This statement is also true. The nucleus, which contains protons and neutrons, is dense and contains most of the atom's mass. **Hint:** Reflect on the composition of the nucleus and its significance in the atom. ### Step 4: Correlate the Statements While both statements are correct, we need to determine if Statement-II provides a valid explanation for Statement-I. The reason why very few alpha particles are deflected is not solely because the nucleus is heavy. Instead, it is primarily due to the fact that most of the atom is empty space. When alpha particles encounter the nucleus, they may be deflected, but since the nucleus occupies a very small volume compared to the entire atom, most particles pass through without hitting it. **Hint:** Consider the spatial distribution of the nucleus relative to the entire atom. ### Step 5: Conclusion Thus, while both statements are true, Statement-II does not correctly explain Statement-I. Therefore, the answer to the question is that both statements are correct, but Statement-II is not the correct explanation for Statement-I. ### Final Answer: Both Statement-I and Statement-II are correct, but Statement-II is not the correct explanation for Statement-I. ---
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Assertion (A) : In rutherford's gold foil experiment, very few alpha particle are deflected back Reason (R ) : Nuclear present inside the atom is heavy

Assertion : In Rutherford's alpha -particle scattering experiment, most of the alpha -particles were deflected by nearly 180^(@) . Reason : The positive charge of the atom is spread throughout the atom that repelled and deflected the positively charged alpha -particles.

Rutherford model: The approximate size of the nucleus can be calculated by using energy conservation theorem in Rutherford's alpha -scattering experiment. If an alpha -particle is projected from infinity with speed v towards the nucleus having Z protons, then the alpha -particle which is reflected back or which is deflected by 180^@ must have approached closest to the nucleus .It can be approximated that alpha particle collides with the nucleus and gets back. Now if we apply the energy conservation equation at initial point and collision point then: (P.E.)_i= 0 , since P.E. of two charge system separated by infinite distance is zero. Finally the particle stops and then starts coming back. 1/2m_alpha v_alpha^2+0=0+(Kq_1q_2)/Rimplies 1/2m_alphav_alpha^2=K(2exxZe)/R implies R=(4KZe^2)/(m_alphav_alpha^2) Thus the radius of nucleus can be calculated using above equation. The nucleus is so small a particle that we can't define a sharp boundary for it Radius of a particular nucleus is calculated by the projection of alpha -particle from infinity at a particular speed. Let this radius is the true radius . If the radius calculation for the same nucleus is made by another alpha -particle with half of the earlier speed, then the percentage error involved in the radius calculation is :

Rutherford model: The approximate size of the nucleus can be calculated by using energy conservation theorem in Rutherford's alpha -scattering experiment. If an alpha -particle is projected from infinity with speed v towards the nucleus having Z protons, then the alpha -particle which is reflected back or which is deflected by 180^@ must have approached closest to the nucleus .It can be approximated that alpha particle collides with the nucleus and gets back. Now if we apply the energy conservation equation at initial point and collision point then: (P.E.)_i= 0 , since P.E. of two charge system separated by infinite distance is zero. Finally the particle stops and then starts coming back. 1/2m_alpha v_alpha^2+0=0+(Kq_1q_2)/Rimplies 1/2m_alphav_alpha^2=K(2exxZe)/R implies R=(4KZe^2)/(m_alphav_alpha^2) Thus the radius of nucleus can be calculated using above equation. The nucleus is so small a particle that we can't define a sharp boundary for it An alpha -particle with initial speed v_0 is projected from infinity and it approaches up to r_0 distance from a nuclie. Then, the initial speed of alpha -particle, which approaches upto 2r_0 distance from the nucleus is :

Statement -1 : Large angle scattering of alpha particles led to the discovery of atomic nucleus. Statement -2 : Entire positive charge of atom is concentrated in the central core.

Statement-I : No two electrons in an atom can have the same values of four quantum number. Because Statement-II : No two electrons in an atom can be simultaneously in the same shell, same subshell, same orbitals and have same spin.

Assertion : During alpha -particles scattering experiment, 99% of the alpha -particles get deflected through very large angles. Reason : The nuclei of atoms are positively charged and repel alpha -particles which are also positively charged.

In the gold foil experiment of Geiger and Marsden, that paved the way for Rutherford's model of an atom, ~ 1.00% of the alpha -particles were found to deflect at angles gt 50^(@) . If one mole of alpha -particles were bombarded on the gold foil, compute the number of alpha -particles that would deflect at angles less than 50^(@) .

Assertion: In Rutherford's experiment, a-particles from a sodium source were allowed to fall on a 104 mm thick gold foil. Most of the particles passed straight through the foil. Reason : The entire positive charge and nearly whole of the mass of an atom is concentrated in the nucleus.

Which of the following statement about Rutherford's model of atom are correct? (i) Considered the nucleus as positively charged. (ii) Established that the alpha -particles are four times as heavy as a hydrogen atom. (iii) Can be compared to solar system. (iv) Was in agreement with Thomson's model.

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