Question

If the wavelength for an electron emitted from $H$-atom is $3.3 \times 10^{-10} m$, then energy absorbed by the electron in its ground state compared to minimum energy required for its escape from the atom, is times. (Nearest integer).
[Given : $h =6.626 \times 10^{-34} \,Js$, Mass of electron $=9.1 \times 10^{-31}$ ]

Answer 1

$ \lambda=\frac{ h }{\sqrt{2 mK }} $
$K =\frac{ h ^2}{2 m \lambda^2} $
$K =\frac{ h ^2}{2 m \lambda^2}=\frac{43.9 \times 10^{-68}}{2 \times 9.1 \times 10^{-31} \times 10.89 \times 10^{-20}}$
$ K =2.215 \times 10^{-18} $
$ E _{ abs }= E _{\text {req }}+ K $
$ \frac{ E _{\text {abs }}}{ E _{\text {req }}}=1+\frac{ K }{ E _{\text {req }}}=1+\frac{2.215 \times 10^{-18}}{13.6 \times 1.602 \times 10^{-19}}=2.0166$