A) hydrogen atom
B) \[H{{e}^{+}}\]
C) \[L{{i}^{2+}}\]
D) \[B{{e}^{3+}}\]
Correct Answer: D
Solution :
For hydrogen like atom, the radius of nth orbit \[r_{n}^{Z}=\frac{{{n}^{2}}}{Z}{{a}_{0}}\] Here, \[{{a}_{0}}=0.51\times {{10}^{-10}}m\] \[\therefore \] \[r_{n}^{Z}=\frac{0.51\times {{10}^{-10}}}{4}m\] In the ground state, \[n=1\] \[\therefore \] \[\frac{0.51\times {{10}^{-10}}}{4}=\frac{{{1}^{2}}}{Z}\times 0.51\times {{10}^{-10}}\] \[\therefore \] \[Z=4\] So, the atom is triply ionised beryllium\[(B{{e}^{3+}})\].
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