A) H-atom
B) \[H{{e}^{+}}\]
C) \[l{{i}^{2+}}\]
D) \[B{{e}^{3+}}\]
Correct Answer: D
Solution :
Radius of hydrogen like atom \[\Rightarrow \] \[r_{n}^{Z}=\frac{{{n}^{2}}}{Z}{{r}_{0}}\] where, \[{{r}_{0}}=0.51\times {{10}^{-10}}m\] and \[r_{n}^{Z}=\frac{0.51\times {{10}^{-10}}}{4}m\] At ground state n = 1, therefore Z = 4 Hence, it is berillium \[(B{{e}^{3+}}),\,N=\frac{n\,(n-1)}{2}\]
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