Question

In Rutherford's scattering experiment an a-particle is projected with K.E. = 4 MeV, minimum distance upto which this a-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. 1.04 × 10^–15 m

Answer 1

Answer: (A)

2.08 × 10^–14 m

Answer 2

$R = \frac{1}{4\pi \in_{0}} \times \frac{2Ze^{2}}{K}$

= 9 × 10⁹ × $\frac{2 \times 29 \times (1.6 \times 10^{–19})^{2}}{4 \times 10^{6} \times 1.6 \times 10^{- 19}}m$

= $\frac{9 \times 2 \times 29 \times 1.6 \times 10^{–16}}{4}m$

= 2.08 × 10^–14 m