An energy of `68eV` is required to excite a hydrogen like atom from its second Bohr orbit to the 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 radiation required to eject the electrons from the first Bohr orbit to infinity.

Answer

`6;489.eV,24.28Å`

Explanation

`DeltaE=13.6 Z^(2)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
`68=13.6 Z^(2)[(1)/(2^(2))-(1)/(3^(2))]`
`Z=6`
`KE=13.6Z^(2)/(n^(2))13.6xx(36)/(1)=489.6 eV`
`E=(hc)/(elambda)(eV)`
`lambda=(hc)/(489.6xxe)25.28`