The de Broglie wavelength of electron of He+ ion is 3.32 Å. If the photon emitted upon de-excitation

Question

The de Broglie wavelength of electron of He+ ion is 3.32 Å. If the photon emitted upon de-excitation of this He+ ion is made to hit H-atom in its ground state so as to liberate electron from it, then what will be the de Broglie wavelength of photoelectron?

Accepted Answer

$\lambda = 3.32.\frac{n}{z}\overset{o}{A} \Rightarrow 3.32 = 3.32 \times \frac{n}{2} \Rightarrow n = 2$

Energy of photon liberated in 2 → 1 transition, $\Delta E = 13.6 \times 2^{2} \times \left( \frac{1}{1^{2}} - \frac{1}{2^{2}} \right) = 40.8\text{ eV}$

∴ K.E. of emitted electron from H-atom = 40.8 – 13.6 = 27.2 eV

Hence, its de Broglie wavelength is given by, $\lambda = \sqrt{\frac{150}{27.2}} = 2.348\overset{o}{A}$

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