Question

The de Broglie wavelength of electron of $H{e}^{+}$ ion is $3.32\AA$. If the photon emitted upon de-excitation of this $H{e}^{+}$ ion is made to hit $H$-atom in its ground state so as to liberate electron from it, then the de Broglie wavelength of photoelectron is:

• A. $2.348\AA$
• B. $1.917\AA$
• C. $3.329\AA$
• D. $3.329pm$

Answer 1

The correct option is A $2.348\AA$

Energy of the photon liberated:
Using formula $E=\frac{hc}{\lambda }$
Where $h$ = planck’s constant $6.626\times {10}^{-34}Js$
$c$ = velocity of light = $3\times {10}^{8}m{s}^{-1}$
and $\lambda$ is the wavelength (given) = $3.32\AA =3.32\times {10}^{-10}m$
Putting all these values in the formula:
$E=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{3.32\times {10}^{-10}}$

= $5.947\times {10}^{-16}J$
Now using formula for de-Broglie wavelength i.e., $\lambda =\frac{h}{\sqrt{m\times E}}$
$m$ = mass of electron = $9.109\times {10}^{-31}kg$
$\lambda =\frac{6.626\times {10}^{-34}}{\sqrt{(9.109\times {10}^{-31}\times 5.947\times {10}^{-16}}})$

= $\frac{6.626\times {10}^{-34}}{5.4171\times {10}^{-46}}$

= $2.348\AA$