Atoms & Nuclei

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46 of 239

Question 46 of 239

The radius of 2nd{2}^{nd} orbit of He+{He}^{+} of Bohr's model is r1{r}_{1} and that of fourth orbit of Be3+{Be}^{3+} is represented as r2.{r}_{2}. Now the ratio r2r1\dfrac{{r}_{2}}{{r}_{1}} is x:1.x:1. The value of xx is _____.

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Answer: 2

The formula to calculate the radius (r)\left(r\right) of any nth{n}^{th} Bohr's orbit is given by
r=r0n2Z...(1)r={r}_{0}\dfrac{{n}^{2}}{Z}...\left(1\right)
where,
r0{r}_{0} is Bohr's radius, nn is the principal quantum number and ZZ is the atomic number.
Thus, for Helium ion,

r1=r0222=2r0...(2)\begin{matrix}{r}_{1} & = & {r}_{0}\dfrac{{2}^{2}}{2} \\ & = & 2{r}_{0}...\left(2\right)\end{matrix}
And, for the Beryllium ion,

r2=r0424=4r0...(3)\begin{matrix}{r}_{2} & = & {r}_{0}\dfrac{{4}^{2}}{4} \\ & = & 4{r}_{0}...\left(3\right)\end{matrix}
Divide equation (3) by equation (2) to obtain the required ratio.

r2r1=4r02r0=2\begin{matrix}\dfrac{{r}_{2}}{{r}_{1}} & = & \dfrac{4{r}_{0}}{2{r}_{0}} \\ & = & 2\end{matrix}
Hence,
x=2x=2.

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