A) 1.4 kJ
B) \[1.4\times {{10}^{4}}\,kJ\]
C) \[1.4\times {{10}^{2}}\,kJ\]
D) \[1.4\times {{10}^{3}}\,kJ\]
Correct Answer: D
Solution :
Key Idea: The ionisation energy is the amount of energy required to take out most loosely bonded electron fromisolated gaseous atom \[\therefore \] lonisation energy of nitrogen = energy of photon \[=N\,\,h\,\frac{c}{\lambda }\] Where \[N=6.02\times {{10}^{23}}\] \[c=3\times {{10}^{8}}\] \[\lambda =854\,\overset{o}{\mathop{A}}\,=854\times {{10}^{-10}}M\] \[=\frac{6.02\times {{10}^{23}}\times 6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{854\times {{10}^{-10}}}\] \[=1.4\times {{10}^{6}}\,J\,\,mo{{l}^{-1}}\] \[=1.4\times {{10}^{3}}\,kJ\,\,mo{{l}^{-1}}\]
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