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Ionization Enthalpy

23.Alpha part...

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23.Alpha particle of 6MeV energy is scattered back from a silver foil. Calculate the maximum volume in which the entire charge of the atom is supposed to be concentrated?

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α -

particles of 6 MeV energy is scattered back from a silver foil. Calculate the maximum volume where the entire charge of the atom is supposed to be accumulated.

Z

= 47

Q. When a silver foil (Z = 47) was used in a

α

-ray scattering experiment, the number of

α

particles scattered at

30^{\circ}

was found to be 200 per minute. If the silver foil is replaced by aluminium (z = 13) foil of same thickness, the number of

α

-particles scattered per minute at

30^{\circ}

is nearly equal to

Q. When a silver foil (Z=47) was used in an

α

-ray scattering experiment, the number of

α

partied scattered at

30^{\circ}

aws found to be 200 per minute. If the silver foil is replaced by aluminium (z=13) foil of same thickness, the number of

α

-particles scattered per minute at

30^{\circ}

is nearly equal to

Q. Observation:
In the alpha particle scattering experiment, it was observed that few of the alpha particles were deflected at angles greater than

90^{\circ}

Inference:
Only a large proportion of the entire charge should be concentrated at the center of the atom and not the mass.

Q. An alpha particle after passing through a potential difference of

2 \times 10^{6} v o l t

falls on a silver foil. The atomic number of silver is

47

(I)

K . E

of the alpha-particle at the time of falling on the foil.

(I I) K . E

of the alpha-particle at a distance of

5 \times 10^{- 14} m

from the nucleus,

(I I I)

the shortest distance from the nucleus of silver to which the alpha-particle reaches.

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