If the wavelength for an electron emitted from H - atom is 3 . 3 ×...

De Broglie's equation for calculating the wave length$( \lambda )$~associated with the electron is given below

$\lambda = \dfrac{h}{p}= \dfrac{h}{mV}$

Let Kinetic energy of the particle of mass 'm' is K

$K = \dfrac{1}{2}{mV}^{2}2Km = m^{2}V^{2}\sqrt{2Km} = mV$

$\lambda = \dfrac{h}{2mK}$

$h$=Planck's constant=$6 . 62 \times {10}^{- 27}erg - sec$

$K = \dfrac{h2}{2{mλ}^{2}}$

$K = \dfrac{h2}{2{mλ}^{2}} = \dfrac{43 . 9 \times {10}^{- 68}}{2 \times 9 . 1 \times {10}^{- 31} \times 10 . 89 \times {10}^{- 20}}$

$K = 2 . 215 \times {10}^{- 18}$

${E}_{abs} = {E}_{req} + K{E}_{abs} = Energyabsorbed{E}_{req =}EnergyrequiredK = Kineticenergy\dfrac{Eabs}{Ereq} = 1 + \dfrac{K}{Ereq} = 1 + \dfrac{2 . 215 \times {10}^{- 18}}{13 . 6 \times 1 . 602 \times {10}^{- 19}} = 2 . 0166$