In Rutherford's scattering experiment, an α -particle is projected with KE =4 MeV , minimum distance up to which this α -particle can reach to the nucleus of Cu nucleus: (atomic number of Cu =29 )

Question

In Rutherford's scattering experiment, an α -particle is projected with KE =4 MeV , minimum distance up to which this α -particle can reach to the nucleus of Cu nucleus: (atomic number of Cu =29 ) (A) 2.08 × 10-14 m (B) 2.08 × 10-15 m (C) 1.04 × 10-14 m (D) 0.04 × 10-15 m

Tardigrade Admin · Accepted Answer

R=(1/4 π ϵ0) × (2 Z e2/K) =9 × 109 × (2 × 29 ×(1.6 × 10-19)2/4 × 106 × 1.6 × 10-19) m =(9 × 2 × 29 × 1.6 × 10-16/4) m =2.08 × 10-14 m

Q. In Rutherford's scattering experiment, an $α$ -particle is projected with $K E = 4 M e V,$ minimum distance up to which this $α$ -particle can reach to the nucleus of $C u$ nucleus: (atomic number of $C u = 29$ )

$2.08 \times 1 0^{- 14} m$

$2.08 \times 1 0^{- 15} m$

$1.04 \times 1 0^{- 14} m$

$0.04 \times 1 0^{- 15} m$

$R = \frac{1}{4 π ϵ _{0}} \times \frac{2 Z e ^{2}}{K}$

$= 9 \times 1 0^{9} \times \frac{2 \times 29 \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{4 \times 1 0 ^{6} \times 1.6 \times 1 0 ^{- 19}} m$

$= \frac{9 \times 2 \times 29 \times 1.6 \times 1 0 ^{- 16}}{4} m$

$= 2.08 \times 1 0^{- 14} m$

Q. In Rutherford's scattering experiment, an α -particle is projected with KE=4MeV, minimum distance up to which this α -particle can reach to the nucleus of Cu nucleus: (atomic number of Cu=29 )

2.08×10−14m

2.08×10−15m

1.04×10−14m

0.04×10−15m