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

Answer: 2

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