Question

13.6 eV is needed for 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?
$\left(m_{e}=9.109\times10^{-31} Kg, e=1.602\times10^{-19} Coloumb, h=6.63\times10^{-34} Js\right)$

• A. $1.5\times10^{-6}m$
• B. $2.3\times10^{-11}m$
• C. $4.7\times10^{-10} m$
• D. $1.4\times10^{-9}m$

Answer 1

Answer: (C)

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 to kinetic energy. KE = 6.8 eV = 6.8 (1.602 × $10^{-19}$ coulomb) (1 volt) = 1.09 × $10^{-18} J$
Now, KE =$\frac{1}{2}m \upsilon^{2}$
$or, \, \upsilon=\sqrt{\frac{2 KE}{m}}=\sqrt{\frac{2\left(1.09\times10^{-18} J\right)}{\left(9.109\times10^{-31}Kg\right)}}=1.55\times10^{6} m/s$
$\therefore \, \lambda=\frac{h}{m \upsilon}=\frac{\left(6.63\times10^{-34} J s\right)}{\left(9.109\times10^{-31}Kg\right)\left(1.55\times10^{6} m/s \right)}=4.7\times10^{-10} m$