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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`)

A

`3.6 xx 10^(-28)m^(3)`

B

`5.97 xx 10^(-42)m^(3)`

C

`6.55 xx 10^(-71)m^(3)`

D

`48 xx 10^(-42)m^(3)`

Text Solution

Verified by Experts

The correct Answer is:
D
`KE = PE`
`(1)/(2) mv^(2) = (1)/(4pi in_(0)).(theta_(1)theta_(2))/(R)`
`r = 2.25 xx 10^(14)m, V = (4)/(3) pi r^(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?

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