Question

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

• A. $2.08 \times 10^{-14} \,m$
• B. $2.08 \times 10^{-15} \,m$
• C. $1.04 \times 10^{-14}\, m$
• D. $0.04 \times 10^{-15} \,m$

Answer 1

Answer: (A)

$R=\frac{1}{4 \pi \epsilon_{0}} \times \frac{2 Z e^{2}}{K}$

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

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

$=2.08 \times 10^{-14} \,m$