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Structure of Atom — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryStructure of Atom3 Marks
Question
The electron energy in hydrogen atom is given by $\text{E}_{\text{n}}=\frac{ (–2.18 \times 10–18 )}{\text{n}^2}\text{J}.$ Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
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Answer

$\text{E}_{\text{n}}=\frac{ (–2.18 \times 10–18 )}{\text{n}^2}\text{J}$
Energy required for ionization from n = 2 is given by,
$\triangle\text{E = E}_{\infty}-\text{E}_2$
$=\Big[\Big(\frac{-2.18\times10^{-18}}{(\infty)^2}\Big)-\Big(\frac{-2.18\times10^{-18}}{(2)^2}\Big)\Big]\text{J}$
$=\Big[\frac{2.18\times10^{-18}}{4}-0\Big]\text{J}$
$=0.545\times10^{-18}\text{J}$
$\triangle\text{E}=5.45\times10^{-19}\text{J}$
$\lambda=\frac{\text{hc}}{\triangle\text{E}}$
Here, $\lambda$ is the longest wavelength causing the transition.
$\lambda=\frac{(6.626\times10^{-34})(3\times10^8)}{5.45\times10^{-19}}=3.647\times10^{-7}\text{m}$
$=3647\times10^{-19}\text{m}$
$=3647\mathring{\text{A}}$

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