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# A singly ionized helium ion has only one electron and is denoted $\mathrm{He}^{+}.$ What is the ion’s radius in the ground state compared to the Bohr radius of hydrogen atom?

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$\textbf{Concept: }$

As we know atom is the smallest unit of elements which is combine to form molecules and the molecules are combining to form the matter. Also, we know that the first direct observation of atoms was in brownian motion which give accurate sizes for atoms in range $10^{-10} \mathrm{~m}$. Also, we know that the atom consists of electrons, protons and neutrons. We know that the charge-to-mass ratios are given by

\begin{align*} \dfrac{q_{e} }{ m_{e} } &= - 1.76 \times 10^{11} \mathrm{~C/kg} \\ \dfrac{ q_{p} }{ m_{p} } &= 9.57 \times 10^{7} \mathrm{~C/kg} \\ \end{align*}

Where, the mass of the electron is $m_{e} = 9.11 \times 10^{-31} \mathrm{~kg}$ and the mass of proton is $m_{p} = 1.67 \times 10^{-27} \mathrm{~kg}$.

According to Bohr's theory of Hydrogen atom, atomic and molecular spectra are quantized with hydrogen spectrum wavelength which is given by

\begin{align*} \dfrac{1}{ \lambda } &= R ~ \left( \dfrac{1}{ n_{f}^{2} } - \dfrac{1}{ n_{i}^{2} } \right) \end{align*}

Where, $\lambda$ is the wavelength of the emitted electromagnetic radiation and $R = 1.097 \times 10^{7} \mathrm{~m^{-1}}$ is the Rydberg constant.

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