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alpha- particles of 6 MeV energy is scat...

`alpha-` particles of `6 MeV` energy is scattered back form a silver foil. Calculate the maximum volume in which the entire positive charge fo the atom is supposed to be concentrated. (`Z` for silver `= 47`)

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The correct Answer is:
`48xx10^(-42) m^(3)`
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An alpha particle of energy 4MeV is scattered through an angle of 180^(@) by a gold foil (Z=79). Calculate the maximum volume in which positive charge of the atom is likely to be concetrated?

Large angle scattering of alpha- particle led Rutherford to the discovery of atomic nucleus. It is the tiny central core of every atom in which entire positive charge of the atom is concentrated. The distance of closet approach of alpha particle from the nucleus is r_(theta)=((Ze)(2e))/(4pi epsilon_(theta)((1)/(2)mv^(2))) This distance r_(theta) gave him the order of size of nucleus. Read the above passage and answer the followign question: (i) What is the distance of closet approach of an alpha particle of energy 7.7 MeV from gold nucleus (z=79)? (ii) What is the implication of this relation in day to day life?

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.

An alpha particle of K.E. 10^(-12)J exibits back scattering form a gold nucleus Z=79. What can be the maximum possible radius of the gold nucleus?

Rutherford's alpha -particle scattering experiment showed that (i) electrons have negative charge. (ii) the mass and positive charge of the atom is concentrated in the nucleus. (iii) neutron exists in the nucleus. (iv) most of the space in atom is empty. Which of the above statements are correct ?

An alpha -particle of velocity 2.3xx10^7 m/s exhibits back scattering by a gold foil (Z=79).Predict the maximum possible radius of the gold nucleus approximately Charge of mass ratio for alpha -particle is 4.8xx10^7 C/kg.

A narrow beam of alpha particles falls normally on a silver foil behind which a counter is set to register the scattered particles. On substraction of platinum foil of the same mass thickness for the silver foil, the number of alpha particles registered per unit time increased eta= 1.52 time s . Find the atomic number of platinum, assuming the atomic number of silver and the atomic masses of both platinum and silver to be known.

Assertion (A) : In Rutherford's experiment, alpha particles from a radium source were allowed to fall on a 10^(-4)mm thick gold foil. Most of the particles passed straight through the foil. Reason (R) : The entire positive charge and nearly whole of the mass of an atom is concentrated in the nucleus.

According to Thomson's model, every atom consists of a positively charged sphere of radius 10^(-10)m in which entire mass and positive charge of the atom are uniformly distributed. Inside the sphere, electrons are embedded like seeds in a watermelon. According Rutherford, entire positive charge and mass of the atom are concentrated in a tiny central core of the atom, which is called atomic nucleus. Size of nucleus ~~10^(-15)m . The nucleus contains proton and neutrons. Negatively charged electrons revolve around the nucleus in circular orbits. Large angle scattering of alpha -particle could not be explained by

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