JEE Main 2021 · Chemistry · Atomic Structure · medium

Monochromatic X-ray radiation of wavelength 100 AA ejects an electron from a He^+ ion that is initially in its n=2 state. The de Broglie wavelength…

Numerical answer

Monochromatic X-ray radiation of wavelength

100\ \mathrm{\AA}

ejects an electron from a

\mathrm{He^+}

ion that is initially in its

n=2

state. The de Broglie wavelength of the ejected electron, in picometres to the nearest integer, is (take

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

; energy of H atom ground state

=-13.6\ \mathrm{eV}

; the electron has

\lambda(\mathrm{\AA})=\sqrt{150.4/K}

with

K

in eV)

Show answer and solution

Answer: 117

Energy of

\mathrm{He^+}

n=2

E=-13.6\dfrac{Z^2}{n^2}=-13.6\times\dfrac{4}{4}=-13.6\ \mathrm{eV}

. Photon energy

=\dfrac{12400}{100}=124\ \mathrm{eV}

. Kinetic energy of the ejected electron

K=124-13.6=110.4\ \mathrm{eV}

. De Broglie wavelength:

\lambda=\sqrt{\dfrac{150.4}{110.4}}=\sqrt{1.362}=1.167\ \mathrm{\AA}=116.7\ \mathrm{pm}\approx117\ \mathrm{pm}

. A common mistake is to use

Z=1

n=1

energy, which changes

K

considerably.

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