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) \[\frac{10.39}{1.5\times {{10}^{6}}}m\]
Correct Answer: D
Solution :
lonisation 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 \[=\frac{10.39}{1.5\times {{10}^{6}}}m\]
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