An electron in a hydrogen atom in its ground state absorbs energy equal to ionisation energy of `Li^(+2)`. The wavelength of the emitted electron is :-
(a). `3.32xx10^(10)m`
(b). `1.17 Å`
(c). `2.32xx10^(9)nm`
(d). `3.33"pm"`

Options

A. `3.32xx10^(-10)m`
B. `1.17Å`
C. `2.32xx10^(-9)nm`
D. `3.33"pm"`

Answer

B

Explanation

K.L of electron K.E`=(1)/(2)mv^(2)=9xx13.6xxe`
De-Briglie wavelenght
`lambda=(h)/(mv)=(h)/sqrt(2m.K.E.)=(6.63xx10^(34))/(sqrt2xx9.01xx10^(-31)xx9xx13.6xx1.6xx10^(-19))`
=`1.17xx110^(-10)` `m=1.17 Å`