JEE Advanced 2016 Paper 1 · Chemistry · Atomic Structure · easy

A sample of ground-state Li^2+ ions absorbs photons of energy exactly 108.8 eV (use En=-13.6 Z^2/n^2 eV for hydrogen-like ions). Each excited ion…

Numerical answer

A sample of ground-state

\mathrm{Li^{2+}}

ions absorbs photons of energy exactly

108.8\ \mathrm{eV}

(use

E_n=-13.6\,Z^2/n^2\ \mathrm{eV}

for hydrogen-like ions). Each excited ion then de-excites to the ground state through all possible transitions, and all spectral lines are observed. How many of the observed emission lines have photon energy greater than

60\ \mathrm{eV}

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Answer: 2

For

\mathrm{Li^{2+}}

(

Z=3

E_n=-\dfrac{13.6\times9}{n^2}=-\dfrac{122.4}{n^2}\ \mathrm{eV}

, so

E_1=-122.4

E_2=-30.6

E_3=-13.6\ \mathrm{eV}

. Absorbing

108.8\ \mathrm{eV}

from the ground state gives a final energy

-122.4+108.8=-13.6\ \mathrm{eV}

, which is exactly

E_3

; hence the ions reach

n=3

(a photon of any other energy would not be absorbed). From

n=3

three emission lines are possible:

3\to1

with

108.8\ \mathrm{eV}

2\to1

with

-30.6-(-122.4)=91.8\ \mathrm{eV}

, and

3\to2

with

-13.6-(-30.6)=17.0\ \mathrm{eV}

. Only the first two exceed

60\ \mathrm{eV}

, so the number of lines with energy greater than

60\ \mathrm{eV}

2

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