A) \[H\]
B) \[H{{e}^{+}}\]
C) \[{{B}^{4+}}\]
D) \[L{{i}^{2+}}\]
E) \[B{{e}^{2+}}\]
Correct Answer: B
Solution :
\[{{E}_{n}}=\frac{-13.6\times {{(Z)}^{2}}}{{{n}^{2}}}eV\] If \[n=1\] \[{{E}_{n}}=-13.6\,{{Z}^{2}}\] \[\therefore \] \[{{Z}^{2}}=\frac{-\,54.4}{-13.6}\] \[{{Z}^{2}}=4\] \[Z=2\] Thus, the species with atomic number 2 (i.e.,\[H{{e}^{+}}\]) has the ionisation energy 54.4 eV.
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