Question

The ionization potential of mercury is 10.38 volt. To gain energy sufficient enough to ionize mercury, an electron must travel in an electric field of 1.5×10⁶Vm^–1 a distance of-

• A. $\frac{10.39}{1.5 \times (10)^{6}}m$
• B. 10.39 × 1.5 × 10⁶ m
• C. 10.39 × 1.6 ×10^–19m
• D. $\frac{10.39 \times 1.6 \times 10^{–19}}{1.5 \times 10^{6}}$m

Answer 1

Answer: (A)

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

Answer 2

E = $\frac{\Delta V}{\Delta r}$

DV = E Dr

10.39 = 1.5 × 10⁶ × Dr

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