Question

The ionisation potential of mercury is 10.39 V. How far an electron must travel in an electric field of $1.5\times {{10}^{6}}V{{m}^{-1}}$ to gain sufficient energy to ionise mercury?
A. $\dfrac{10.39}{1.6\times {{10}^{-19}}}m$
B. $\dfrac{10.39}{2\times 1.6\times {{10}^{-19}}}m$
C. $10.39\times 1.6\times {{10}^{-19}}m$
D. $\dfrac{10.39}{1.5\times {{10}^{6}}}m$

Answer 1

Hint: Ionisation potential is equal to the ionisation energy divided by the charge of an electron. When the electron travels in the electric field it will accelerate and gain some energy. Use the formula for the energy gained by the electron. Equate the ionisation energy of mercury and energy gained by the electron.

Formula used:
$V=\dfrac{U}{e}$,
where $V$ is ionisation potential, $U$ is ionisation energy and $e$ is the magnitude of charge on an electron.
$W=eEd$,
where $W$ is the energy gained by an electron when it travels a distance d along an electric field E.

Complete step by step answer:
 Ionisation potential energy of a substance is the minimum required energy required to ionise the substance. Ionisation potential is equal to the ionisation energy divided by the charge of an electron.
$V=\dfrac{U}{e}$ …. (i)
It is given that the ionisation potential of mercury is 10.39V. We are asked to calculate the distance travelled by an electron in an electric field of $1.5\times {{10}^{6}}V{{m}^{-1}}$ such that it attains the minimum energy to ionise mercury. This means that $V=10.39volts$.
And $e=1.6\times {{10}^{-19}}C$.
Substitute the values of $e$ and $V$ in (i)
$10.39=\dfrac{U}{1.6\times {{10}^{-19}}}$
$\Rightarrow U=10.39\times 1.6\times {{10}^{-19}}J$ …. (ii)
When the electron travels in the electric field it will accelerate and gain some energy. The energy gained by the electron is equal to $W=eEd$.
Here, $E=1.5\times {{10}^{6}}V{{m}^{-1}}$.
$\Rightarrow W=1.6\times {{10}^{-19}}\times 1.5\times {{10}^{6}}d$ ….. (iii)
And to ionise the mercury, the energy gained by the electron must be equal to the ionisation energy of mercury i.e, $W=U$.
Therefore, equate (ii) and (iii)
$10.39\times 1.6\times {{10}^{-19}}=1.6\times {{10}^{-19}}\times 1.5\times {{10}^{6}}d$
$\therefore d=\dfrac{10.39}{1.5\times {{10}^{6}}}$.

Hence, the correct option is D.

Note: Sometimes it is misunderstood by thinking that ionisation potential and ionisation energy are one and the same quantities. However, they are not the same quantities and we have seen the difference between the two.We can say that ionisation potential is the minimum amount of potential difference across a substance required to ionise the substance.