Question

The ionisation potential of mercury is $10.39\, V$. How far ah electron must travel in an electric field of $ 1.5\times {{10}^{6}}V/m $ to gain sufficient energy to ionise mercury?

• A. $ \frac{10.39}{1.6\times {{10}^{-19}}}m $
• B. $ \frac{10.39}{2\times 1.6\times {{10}^{-19}}}m $
• C. $ 10.39\times 1.6\times {{10}^{-19}}m $
• D. $ \frac{10.39}{1.5\times {{10}^{6}}}m $

Answer 1

Answer: (D)

Ionisation potential $(V)$ of mercury is the energy required to strip it of an electron. The electric field strength is given by
$E=\frac{V}{d}$
where, $d$ is distance between plates creating electric field.
Given, $V=10.39\, V , E=1.5 \times 10^{6} V / m$
$\therefore d=\frac{V}{E}=\frac{10.39}{1.5 \times 10^{6}} m$
Hence, distance travelled by electron to gain ionization energy is
$=\frac{10.39}{1.5 \times 10^{6}} m$