Question

An energy of 68.0 eV is required to excite a hydrogen-like atom in its second Bohr orbit to third. The nuclear charge is Ze. Find the value of Z, the kinetic energy of the electron in the first Bohr orbit and the wavelength of the electronmagnetic radiation required to eject the electron from the first orbit to infinity.

Answer 1

`n_(1)=2,n_(2)=3m Delta E =68 eV`
`Delta E=13.6 Z^(2)((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`
`68=13.6Z^(2)((1)/(2^(2))-(1)/(3^(2))=(1)/(4)-(1)/(9)=(9-4)/(36)=(5)/(36))`
`Z^(2)=(68xx36)/(13.6xx5)=36`
Z=6
`K_(n)=-K_(n)=-(-(13.6Z^(2))/(n^(2)))=(13.6Z^(2))/(n^(2))`
For first Bohr orbit, n=1
`K_(1)=(13.6xx6^(2))/(1^(2))=489.6 eV`
`(1)/(lambda)=Z^(2)R((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`
`(6)^(2) R((1)/(1^(2))-(1)/(oo))=36 R`
` lambda=(1)/(36R) =(1)/(36xx1.097xx10^(7))=0.025xx10^(-7) m`
`=2.5xx10^(9) m=2.5 nm`