An electron in hydrogen like atom is in excited state 2t has a total energy of -3.4eV. Calculate (a) the kinetic energy (b) the de-Broglie wavelength of the electron. [Given `h=6.63xx10^(-34)Js`]

Explanation

(a) Total energy of electron in a hydrogen like atom is given by
`E=KE+PE`
Also, `KE=1/2(PE) or PE=-2KE`
`:. E=KE-2KE=-KE=(-3.4eV)`
`=3.4eV`
(b) De-Broglie wavelength of electron is
`h=h/(mv)=h/([2m(KE)]^(1//2))`
`=(6.63xx10^(-34))/([2xx(9.1xx10^(-31))xx(3.4xx1.6xx10^(-19))]^(1//2))`
`=(6.63xx10^(-34))/(9.95xx10^(-25))=6.66xx10^(-10)m=6.66Å`