Question

$13.6\text{eV}$ is required for the ionization of a hydrogen atom. An electron in a hydrogen atom in its ground state absorbs $14.83\text{eV}$ energy as the minimum required for it to escape from the atom. What is the wavelength of the emitted electron?

$(\text{Given :}{m}_e=9.1\times {10}^{–31}\text{kg},e=1.6\times {10}^{–19}\text{coulomb},h=6.6\times {10}^{–34}J.s),1.23\times 1.6\approx 1.97,2\times 0.197\times 9.1\approx 3.6$

• A. $1.1\times {10}^{-9}m$
• B. $2.2\times {10}^{-6}m$
• C. $7.40\times {10}^{-10}m$
• D. $1.1\times {10}^{-10}m$

Answer 1

The correct option is A $1.1\times {10}^{-9}m$

$14.83\text{eV}$ is absorbed by the hydrogen atom out of which $1.23\text{eV}(14.83–13.6)$ is converted into kinetic energy.
$K.E.=12.3\text{eV}=1.23(1.6\times {10}^{–19})=0.197\times {10}^{-18}\text{J}$

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

$=\frac{6.6\times {10}^{-34}J.sec}{\sqrt{2\times 0.197\times {10}^{-18}J\times 9.1\times {10}^{-31}Kg}}=\frac{6.6\times {10}^{-34}}{\sqrt{3.6\times {10}^{-49}}}=\frac{6.6\times {10}^{-34}}{6\times {10}^{-25}}=1.1\times {10}^{-9}\text{m}$

The wavelength associated with the electron is $=1.1\times {10}^{-9}\text{m}$