Question

An a - particle after passing through a potential difference of V-volts collides with a nucleus. If the atomic number of the nucleus is Z then the distance of closest approach of a-particle to the nucleus will be-

• A. 14.4 $\frac{Z}{V}$Å
• B. 14.4 $\frac{Z}{V}$m
• C. 14.4 $\frac{Z}{V}$cm
• D. All of these

Answer 1

Answer: (A)

14.4 $\frac{Z}{V}$Å

Answer 2

K.E. = P.E. = qV

2eV = $\frac{K(Ze)(2e)}{d}$ = qV

\ d = $\frac{9 \times 10^{9} \times Z \times e \times 2e}{2eV}$

\ d = $\frac{9 \times 10^{9} \times 1.6 \times 10^{–19} \times Z}{V}$

d = 14.4 × 10^–10 $\left( \frac{Z}{V} \right)$m