JEE Main 2018 · Chemistry · Atomic Structure · hard

Light of wavelength 200 AA (take hc=12400 eV AA) is absorbed by a Li^2+ ion that is initially in its first excited state. The ejected electron has a…

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Light of wavelength

200\ \text{\AA}

(take

hc=12400\ \mathrm{eV\,\AA}

) is absorbed by a

\mathrm{Li^{2+}}

ion that is initially in its first excited state. The ejected electron has a de Broglie wavelength of about (use

\lambda=12.27/\sqrt{K}\ \text{\AA}

with

K

in eV; ground-state energy of hydrogen

=-13.6

eV)

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Answer: (C) $2.19\ \text{\AA}$

Photon energy

=12400/200=62

eV. First excited state of

\mathrm{Li^{2+}}

(

Z=3

n=2

E=-13.6\times\dfrac{9}{4}=-30.6

eV, so the binding energy is

30.6

eV. Kinetic energy of the electron

=62-30.6=31.4

eV.

\lambda=12.27/\sqrt{31.4}=2.19\ \text{\AA}

. Ignoring

Z^2

(binding

3.4

eV) gives

K=58.6

eV and

1.60\ \text{\AA}

; ignoring binding altogether gives

1.56\ \text{\AA}

Original question written in the JEE Main 2018 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake