A) 8 Å
B) 4 Å
C) 0.5 Å
D) 0.25 Å
Correct Answer: B
Solution :
By using \[{{r}_{n}}={{r}_{0}}\frac{{{n}^{2}}}{Z}\]; Where r0 = Radius of the Bohr orbit in the ground state atom . So for \[H{{e}^{+}}\]third excited state \[n=4,\ Z=2,\ {{r}_{0}}=0.5{\AA}\]Þ \[{{r}_{4}}=0.5\times \frac{{{4}^{2}}}{2}=4{\AA}\]
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