[Four-digit Integer] An α-particle of momentum 3.2 × 10^-20 kg ms^-1 is projected towards the nucleus of an atom of an element. If the distance of closest approach of α-particle is 1.5 × 10^-13 m, the

Question

[Four-digit Integer] An α-particle of momentum 3.2 × 10^-20 kg ms^-1 is projected towards the nucleus of an atom of an element. If the distance of closest approach of α-particle is 1.5 × 10^-13 m, the atomic number of element is (Mass of α-particle = 4 amu, charge on electron = 1.6 × 10^-19 coulomb, N_A = 6 × 10^23)

Prashant Kotian · Accepted Answer

Correct answer: 0025. K.E. = P^2/(2m) = (3.2×10^-20)^2 / (2×6.67×10^-27) = 7.68×10^-14 J. At closest approach: K.E. = 9×10^9 × 2Z × (1.6×10^-19)^2 / (1.5×10^-13) = 3.072×10^-15 × Z. Z = 7.68×10^-14 / 3.072×10^-15 = 25.

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