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)

$\frac{10.39}{1.5\times {{10}^{6}}}m$

Answer 2

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 $=\frac{10.39}{1.5 \times 10^{6}} m$