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JEE MAIN 2023
13-04-23 S1
Question
The radius of $2^{\text {nd }}$ orbit of $\mathrm{He}^{+}$of Bohr's model is $\mathrm{r}_1$ and that of fourth orbit of Be ${ }^{3+}$ is represented as $\mathrm{r}_2$. Now the ratio $\frac{r_2}{r_1}$ is $x$: 1. The value of $x$ is $\_\_\_\_$
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JEE Main
Physics
Easy
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