In Geiger-Marsden experiment, in a head-on collision between an α-particle and a gold nucleus, the minimum separation is 4 × 10-14 m. The energy of α-particle is (Take, atomic number of gold =79 )

Question

In Geiger-Marsden experiment, in a head-on collision between an α-particle and a gold nucleus, the minimum separation is 4 × 10-14 m. The energy of α-particle is (Take, atomic number of gold =79 ) (A) 5.68 MeV (B) 8 MeV (C) 4.47 MeV (D) 7.24 MeV

Tardigrade Admin · Accepted Answer

At distance of minimum separation, KE = PE. ⇒ KE =(k Ze ⋅ 2 e/r) or KE =(9 × 109 × 79 × 2 ×(1.6 × 10-19)2/4 × 10-14 J ) =(9 × 109 × 79 × 2(1.6 × 10-19)2/4 × 10-14 × 1.6 × 10-19) eV =5.688 MeV (∵ 1 MeV =1.6 × 10-13 J )

Q. In Geiger-Marsden experiment, in a head-on collision between an α-particle and a gold nucleus, the minimum separation is 4×10−14m. The energy of α-particle is (Take, atomic number of gold =79 )

5.68 MeV

8 MeV

4.47 MeV

7.24 MeV

At distance of minimum separation, KE=PE. ⇒KE=rkZe⋅2e​ or KE=4×10−14J9×109×79×2×(1.6×10−19)2​ =4×10−14×1.6×10−199×109×79×2(1.6×10−19)2​eV =5.688MeV(∵1MeV=1.6×10−13J)

Q. In Geiger-Marsden experiment, in a head-on collision between an $α$ -particle and a gold nucleus, the minimum separation is $4 \times 1 0^{- 14} m$ . The energy of $α$ -particle is (Take, atomic number of gold $= 79$ )