Question

13.6 eV is required for the ionization of a hydrogen atom. 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?

$(m_e=9.109\times {10}^{–31}kg,e=1.602\times {10}^{–19}$ coulomb, $h=6.63\times {10}^{–34}J.s)$

• A. $4.70\times {10}^{-10}m$
• B. $1.55\times {10}^{-6}m$
• C. $7.40\times {10}^{-10}m$
• D. $5.50\times {10}^{-6}m$

Answer 1

The correct option is A $4.70\times {10}^{-10}m$

1.5 times of 13.6 eV i.e., 20.4 eV is absorbed by the hydrogen atom out of which 6.8 eV (20.4 – 13.6) is converted into kinetic energy.
$K.E.=6.8eV=6.8(1.602\times {10}^{–19})=1.09\times {10}^{-18}J.$

$\lambda =\frac{\lambda }{\sqrt{2\times K.E.\times m_e}}$

$=\frac{6.636\times {10}^{-34}J.sec}{\sqrt{2\times 1.09\times {10}^{-18}J\times 9.109\times {10}^{-31}Kg}}$

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

The wavelength associated with the electron is $=4.70\times {10}^{-10}m$