Q. The ionisation potential of mercury is 10.39 V. How far an electron must travel in an electric field of $1.5\ \times \ 10^{6}\ Vm^{- 1}$ to gain sufficient energy to ionize mercury?
A) $\frac{10.39}{1.5\ \times 10^{6}} \times 1.0\ \times \ 10^{- 19}\ m$
B) $\frac{10.39}{1.5\ \times 10^{6}}$m
C) $1.39\ \times \ 1.6\ \times \ 10^{- 19}\ m$
D) $\frac{10.39}{1.6\ \times 10^{- 19}}$m
Explanation
(b) Electric field E =$\frac{V}{d}$
$d = \frac{V}{E}$
=$\frac{10.39}{1.5 \times 10^{6}}$ $m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.