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JEE Main 2016 · Chemistry · Atomic Structure · hard

Light emitted when a Li^2+ ion de-excites from n=3 to n=2 falls on a metal of work function 2.5 eV. Using En=-13.6 Z^2/n^2 eV, h=6.63×10^-34 J s…

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Light emitted when a Li2+\mathrm{Li^{2+}} ion de-excites from n=3n=3 to n=2n=2 falls on a metal of work function 2.5 eV2.5\ \mathrm{eV}. Using En=−13.6 Z2/n2 eVE_n=-13.6\,Z^2/n^2\ \mathrm{eV}, h=6.63×10−34 J sh=6.63\times10^{-34}\ \mathrm{J\,s}, me=9.1×10−31 kgm_e=9.1\times10^{-31}\ \mathrm{kg} and 1 eV=1.6×10−19 J1\ \mathrm{eV}=1.6\times10^{-19}\ \mathrm{J}, the de Broglie wavelength of the fastest photoelectron is nearly
  1. A. 3.0 A˚3.0\ \text{\AA}
  2. B. 5.5 A˚5.5\ \text{\AA}
  3. C. 3.2 A˚3.2\ \text{\AA}
  4. D. 1.6 A˚1.6\ \text{\AA}
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Answer: (C) 3.2 A˚3.2\ \text{\AA}

For Li2+\mathrm{Li^{2+}}, Z=3Z=3: ΔE=13.6(9)(14−19)=17.0\Delta E=13.6(9)\left(\tfrac14-\tfrac19\right)=17.0 eV. Maximum kinetic energy =17.0−2.5=14.5=17.0-2.5=14.5 eV =2.32×10−18=2.32\times10^{-18} J. λ=h2mK=6.63×10−342(9.1×10−31)(2.32×10−18)=3.2×10−10\lambda=\dfrac{h}{\sqrt{2mK}}=\dfrac{6.63\times10^{-34}}{\sqrt{2(9.1\times10^{-31})(2.32\times10^{-18})}}=3.2\times10^{-10} m =3.2=3.2 Å. Forgetting to subtract the work function gives 3.03.0 Å; using Z=2Z=2 (a He+\mathrm{He^+} slip) gives 5.55.5 Å.
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