Question

$13.6eV$ is needed for ionisation of a hydrogen atom. An electron in a hydrogen atom in its ground state absorbs 1.50 times as much energy as the minimum energy required for it to escape from the atom. If the wavelength of emitted electron is $x\stackrel{o}{A}$, find the nearest integer value of x
$(m_e=9.1\times {10}^{-31}kg,h=6.626\times {10}^{-34}Js)$

Answer 1

Given, ionisation energy = 13.6 eV
Energy absorbed by the electron = $13.6\times 1.5=20.4eV$
So, the maximum kinetic energy of the electron
$=20.4-13.6eV=6.8eV$

Again, from de Broglie wavelength,
$\lambda =\frac{h}{\sqrt{2meV}}$
$\Rightarrow \lambda =\frac{6.626\times {10}^{-34}}{\sqrt{2\times 6.8\times 1.6\times {10}^{-19}\times 9.1\times {10}^{-31}}}$
$\Rightarrow \lambda =\frac{6.626\times {10}^{-34}}{14\times {10}^{-25}}$
$\Rightarrow \lambda =4.73A^o\approx 5\stackrel{o}{A}$