An electron in a hydrogen atom in its ground state absorbs 1 : 50 times as much energy as the minimum required for it to escape from the atom. What is the wavelength of the emitted electron ?

As 13.6 eV is needed for ionization, 20.4 eV  $(13.6\times 1.5)$ must have been absorbed.  Of this, 6.8 eV is converted to kinetic energy.

$\lambda =\frac{h}{\sqrt{(2mK.E)}}$

$=\frac{6.626\times {10}^{34}Js}{\sqrt{(2\times 9.1\times {10}^{31}\times 6.8\times 1.6\times {10}^{19}J)}}$

$=4.70\times {10}^{-10}m=4.70\AA$