Atoms & Nuclei

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Question 78 of 239

A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980A~…980 Ã…. The radius of the atom in the excited state, in terms of Bohr radius a0a_{0}, will be:

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Answer: (C) 16a016 \mathrm{a}_{0}

Energy of photon =hcλ=12500980=12.75eV=\frac{\mathrm{hc}}{\lambda}=\frac{12500}{980}=12.75 \mathrm{eV}

Energy of electron in nth \mathrm{n}^{\text {th }} orbit is given by

En=−13.6n2⇒En−E1=−13.6[1n2−112]\mathrm{En}=\frac{-13.6}{\mathrm{n}^{2}} \Rightarrow \mathrm{E}_{\mathrm{n}}-\mathrm{E}_{1}=-13.6\left[\frac{1}{\mathrm{n}^{2}} \frac{-1}{1^{2}}\right]

⇒12.75=13.6[112−1n2]⇒n=4\Rightarrow 12.75=13.6\left[\frac{1}{1^{2}} \frac{-1}{\mathrm{n}^{2}}\right] \Rightarrow \mathrm{n}=4

∴\therefore Electron will excite to n=4\mathrm{n}=4 We know that ' R′∝n2R^{\prime} \propto n^{2}

∴\therefore Radius of atom will be 16a016 \mathrm{a}_{0}

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