Atoms & Nuclei

JEE Main level, numerical answer, medium

126 of 239

Question 126 of 239

A beam of monochromatic light is used to excite the electron in Li++{Li}^{++} from the first orbit to the third orbit. The wavelength of monochromatic light is found to be x×10−10mx\times {10}^{-10}m. The value of xx is _____ .
[Given hc=1242eVnmhc=1242eVnm]

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Answer: 114

Energy released in the transition will be,
ΔE=13.6×Z2[112−132]\Delta E=13.6\times {Z}^{2}\left[\dfrac{1}{{1}^{2}}-\dfrac{1}{{3}^{2}}\right] =13.6×32×89eV=108.8eV=13.6\times {3}^{2}\times \dfrac{8}{9}eV=108.8eV
Now energy of photon can be written as,
hcλ=108.8eV⇒λ=12400108.8Ao≈114Ao\dfrac{hc}{\lambda }=108.8eV\Rightarrow \lambda =\dfrac{12400}{108.8}A^{o}\approx 114A^{o}

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