MCQ
An $\alpha$-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 $\alpha$-particle to the nucleus will be
- ✓$14.4 \frac{Z}{V} Å$
- B$14.4 \frac{ Z }{ V } m$
- C$14.4 \frac{ Z }{ V } cm$
- D$1.44 \times 10^{-15} m$
