A Li^2+ ion (hydrogen-like, Z=3) in…

A Li^2+ ion (hydrogen-like, Z=3) in the n=3 level makes a transition to n=2 and emits a photon. This photon is absorbed by a hydrogen atom in its…

Numerical answer

\mathrm{Li^{2+}}

ion (hydrogen-like,

Z=3

) in the

n=3

level makes a transition to

n=2

and emits a photon. This photon is absorbed by a hydrogen atom in its ground state and ionises it, the ejected electron carrying away all the excess energy (take the nucleus to be infinitely heavy). Use the hydrogen ground-state energy

-13.6\ \mathrm{eV}

h=6.63\times10^{-34}\ \mathrm{J\,s}

m_e=9.11\times10^{-31}\ \mathrm{kg}

and

e=1.6\times10^{-19}\ \mathrm{C}

. The de Broglie wavelength of the ejected electron, in ångström (

1\ \text{\AA}=10^{-10}\ \mathrm{m}

), is ______. Give the answer correct to two decimal places.

Show answer and solution

Answer: 6.66

For a hydrogen-like ion

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

, so the photon energy is

h\nu=13.6\times9\left(\dfrac14-\dfrac19\right)=13.6\times9\times\dfrac{5}{36}=17.0\ \mathrm{eV}

. Ionising hydrogen from

n=1

needs

13.6\ \mathrm{eV}

, so the electron's kinetic energy is

K=17.0-13.6=3.4\ \mathrm{eV}=5.44\times10^{-19}\ \mathrm{J}

. Its momentum is

p=\sqrt{2mK}=\sqrt{2(9.11\times10^{-31})(5.44\times10^{-19})}=9.956\times10^{-25}\ \mathrm{kg\,m\,s^{-1}}

. The de Broglie wavelength is

\lambda=\dfrac hp=\dfrac{6.63\times10^{-34}}{9.956\times10^{-25}}=6.659\times10^{-10}\ \mathrm{m}\approx6.66\ \text{\AA}

. Using the full

17.0\ \mathrm{eV}

as kinetic energy would incorrectly give

2.98\ \text{\AA}

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