Question

The ionisation potential of mercury is $10.39V$. 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, $V=10.39V,E=1.5\times 10^{6}V/m$
$\therefore d=\frac{V}{E}=\frac{10.39}{1.5\times 10^6}m$
Hence, distance travelled by electron to gain ionization energy is
$=\frac{10.39}{1.5\times 10^6}m$