Atomic Structure Question 47
Question: The radius of which of the following orbit is same as that of the first Bohr’s orbit of hydrogen atom
Options:
A) $ H{e^{+}}(n=2) $
B) $ L{i^{2+}}(n=2) $
C) $ L{i^{2+}}(n=3) $
D) $ B{e^{3+}}(n=2) $
Show Answer
Answer:
Correct Answer: D
Solution:
$ r_{H}=0.529\frac{n^{2}}{z}{\AA} $ For hydrogen ; $ n=1 $ and $ z=1 $ therefore $ r_{H}=0.529{\AA} $ For $ B{e^{3+}}:,Z=4 $ and $ n=2 $ Therefore $ {r_{B{e^{3+}}}}=\frac{0.529\times 2^{2}}{4}=0.529{\AA} $ .
