JEE Mains · Physics · STD 12 - 12. atoms
The radius of \(2^{\text {nd }}\) orbit of \(He ^{+}\)of Bohr's model is \(r _1\) and that of fourth orbit of \(Be ^{3+}\) is represented as \(r _2\). Now the ratio \(\frac{r_2}{r_1}\) is \(x : 1\). The value of \(x\) is .........
- A \(4\)
- B \(6\)
- C \(2\)
- D \(10\)
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
\(r \propto \frac{ n ^2}{ z }\) \(\frac{ r _{ He ^{+}}}{ r _{ Be ^{3+}}}=\frac{2^2 \times 4}{2 \times 4 \times 4}=\frac{1}{2}\)
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