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NEET UG PHYSICS MCQ: ATOMS

1 Hard
1 of 1
Application +4 | -1
An α\alpha-particle after passing through a potential difference of 2×106 V2 \times 10^6 \mathrm{~V} falls on a silver foil. The atomic number of silver is 47. Calculate (i) the kinetic energy of the α\alpha-particle at the time of falling on the foil (ii) the kinetic energy of the α\alpha-particle at a distance of 5×10−14 m5 \times 10^{-14} \mathrm{~m} from the nucleus and (iii) the shortest distance from the nucleus of silver to which the α\alpha-particle reaches.
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Correct option: A
6.4×10−13 J; 2.1×10−13 J; 3.4×10−14 m6.4 \times 10^{-13} \mathrm{~J};\,2.1 \times 10^{-13} \mathrm{~J};\,3.4 \times 10^{-14} \mathrm{~m}
Option A (6.4×10−13 J; 2.1×10−13 J; 3.4×10−14 m6.4 \times 10^{-13} \mathrm{~J};\,2.1 \times 10^{-13} \mathrm{~J};\,3.4 \times 10^{-14} \mathrm{~m}) is correct because it accurately calculates the kinetic energies and shortest distance based on the potential difference and atomic number.

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